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diff --git a/TangDou/AcWing/BeiBao/278.md b/TangDou/AcWing/Bag/278.md
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index 36e5153..719b1e8 100644
--- a/TangDou/AcWing/BeiBao/278.md
+++ b/TangDou/AcWing/Bag/278.md
@@ -131,7 +131,7 @@ int main() {
 $A:$我们可以将结论推广到不同属性的情况下，本题的属性是数量，但如果是最大价值呢？
 我们不难得到需要将$f[0]$初始化为$0$，$f[1\sim n]$初始化为负无穷
 
-为什么要这样设置呢？因为每一个新状态，都需要知道它可以从哪些旧状态转移而来，如果上一个状态是合法的，那么有可能从上一个状态转移而来，但如何标识上一个状态是不是合法呢？比如如果初始化状态值是$0$，并且上一个状态是$0$,表示的是目前的最大值，那是不是合法呢？不好说啊，为什么呢？
+为什么要这样设置呢？因为每一个新状态，都需要知道它可以从哪些旧状态转移而来，如果上一个状态是合法的，那么有可能从上一个状态转移而来，但如何标识上一个状态是不是合法呢？比如如果初始化状态值是$0$，并且上一个状态是$0$,表示的是目前的最大值，那要是不合法呢？不好说啊，为什么呢？
 
 * 上一个状态不合法，没有状态转移过来
 * 上一个状态合法，因为有负数等原因，造成最大值确实为$0$
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diff --git a/TangDou/AcWing/BeiBao/423.md b/TangDou/AcWing/Bag/423.md
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index 4297128..9f70945 100644
--- a/TangDou/AcWing/BeiBao/423.md
+++ b/TangDou/AcWing/Bag/423.md
@@ -37,13 +37,16 @@ $1≤T≤1000$,$1≤M≤100$
 
 
 ### 二、题目解析
-注意一下$m$和$n$的录入顺序。
 
 **01背包模型**
 
-**状态表示$f(i,j)$—集合**： 考虑前 $i$ 个物品，且当前已使用体积为$ j$ 的方案
+**状态表示**
 
-**状态表示$f(i,j)$—属性**： 该方案的价值为最大值 $max$
+$f(i,j)$
+- **集合**
+考虑前 $i$ 个物品，且当前已使用体积为$ j$ 的方案
+- **属性** 
+该方案的价值为最大值 $max$
 
 **状态转移$f(i,j)$**： 
 
@@ -65,7 +68,6 @@ $$f(i,j)=\begin{equation}
 
 
 ### 三、二维朴素作法
-空间复杂度：$O(n×m)$
 时间复杂度：$O(n×m)$
 
 ```cpp {.line-numbers}
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similarity index 100%
rename from TangDou/AcWing/BeiBao/【总结】最小价值-空间恰好j.md
rename to TangDou/AcWing/Bag/【总结】最小价值-空间恰好j.md
diff --git a/TangDou/AcWing/BeiBao/【总结】最小价值-空间至少是j.md b/TangDou/AcWing/Bag/【总结】最小价值-空间至少是j.md
similarity index 100%
rename from TangDou/AcWing/BeiBao/【总结】最小价值-空间至少是j.md
rename to TangDou/AcWing/Bag/【总结】最小价值-空间至少是j.md
diff --git a/TangDou/AcWing/BeiBao/【总结】背包问题的至多_恰好_至少-主帖.md b/TangDou/AcWing/Bag/【总结】背包问题的至多_恰好_至少-主帖.md
similarity index 100%
rename from TangDou/AcWing/BeiBao/【总结】背包问题的至多_恰好_至少-主帖.md
rename to TangDou/AcWing/Bag/【总结】背包问题的至多_恰好_至少-主帖.md
diff --git a/TangDou/AcWing/BeiBao/【总结】脑图形式(一维)总结.md b/TangDou/AcWing/Bag/【总结】脑图形式(一维)总结.md
similarity index 100%
rename from TangDou/AcWing/BeiBao/【总结】脑图形式(一维)总结.md
rename to TangDou/AcWing/Bag/【总结】脑图形式(一维)总结.md
diff --git a/TangDou/AcWing/BeiBao/【总结】脑图形式(二维)总结.md b/TangDou/AcWing/Bag/【总结】脑图形式(二维)总结.md
similarity index 100%
rename from TangDou/AcWing/BeiBao/【总结】脑图形式(二维)总结.md
rename to TangDou/AcWing/Bag/【总结】脑图形式(二维)总结.md
diff --git a/TangDou/AcWing/Bag/背包问题专题,md b/TangDou/AcWing/Bag/背包问题专题,md
new file mode 100644
index 0000000..2e193bd
--- /dev/null
+++ b/TangDou/AcWing/Bag/背包问题专题,md
@@ -0,0 +1,174 @@
+## 背包问题
+
+### 一、$01$背包基础题
+**[$AcWing$ $2$. $01$背包问题](https://www.acwing.com/problem/content/2/)**
+
+**[$AcWing$ $423$. 采药](https://www.acwing.com/problem/content/425/)**
+
+**[$AcWing$ $1024$. 装箱问题](https://www.acwing.com/problem/content/1026/)**
+
+二维状态表示
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+const int M = 1010;
+
+int n, m;
+int w[N], v[N];
+int f[N][M];
+
+int main() {
+    cin >> m >> n;
+
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            f[i][j] = f[i - 1][j]; // 不选
+            if (j >= v[i])
+                f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]); // 选
+        }
+    printf("%d\n", f[n][m]);
+    return 0;
+}
+
+```
+
+一维状态表示
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+
+int n, m;
+int v[N], w[N];
+int f[N];
+
+int main() {
+    cin >> m >> n;
+    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
+
+    // 01背包模板
+    for (int i = 1; i <= n; i++)
+        for (int j = m; j >= v[i]; j--)
+            f[j] = max(f[j], f[j - v[i]] + w[i]);
+
+    printf("%d\n", f[m]);
+    return 0;
+}
+```
+
+
+### 二、二维费用$01$背包问题
+**[$AcWing$ $1022$. 宠物小精灵之收服](https://www.acwing.com/problem/content/1024/)**
+
+**[$AcWing$ $8$. 二维费用的背包问题](https://www.cnblogs.com/littlehb/p/15684961.html)**
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 110;   // 野生小精灵的数量
+const int M1 = 1010; // 小智的精灵球数量
+const int M2 = 510;  // 皮卡丘的体力值
+
+int n, m1, m2;
+int f[M1][M2]; // 一维：精灵球数量，二维：皮卡丘的体力值，值：抓到的小精灵数量最大值
+
+int main() {
+    cin >> m1 >> m2 >> n;
+    m2--; // 留一滴血
+
+    // 二维费用01背包
+    // 降维需要将体积1、体积2倒序枚举
+    for (int i = 1; i <= n; i++) {
+        int v1, v2;
+        cin >> v1 >> v2;
+        for (int j = m1; j >= v1; j--)
+            for (int k = m2; k >= v2; k--)
+                f[j][k] = max(f[j][k], f[j - v1][k - v2] + 1); // 获利就是多了一个小精灵
+    }
+    // 最多收服多少个小精灵[在消耗精灵球、血极限的情况下，肯定抓的是最多的，这不废话吗]
+    printf("%d ", f[m1][m2]);
+
+    // 找到满足最大价值的所有状态里，第二维费用消耗最少的
+    int cost = m2;
+    for (int i = 0; i <= m2; i++) // 如果一个都不收服，则体力消耗最少，消耗值为0
+        if (f[m1][i] == f[m1][m2])
+            cost = min(cost, i);
+
+    // 收服最多个小精灵时皮卡丘的剩余体力值最大是多少
+    printf("%d\n", m2 + 1 - cost);
+    return 0;
+}
+```
+
+**总结**
+- $01$背包，还是背一维的形式比较好，一来代码更短，二来空间更省，倒序就完了。
+- 二维费用的$01$背包，简化版本的$01$背包模板就有了用武之地，因为三维数组可能会爆内存。
+
+### 三、$01$背包之恰好装满
+**[$AcWing$ $278$. 数字组合](https://www.acwing.com/problem/content/280/)**
+
+**二维代码**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+const int M = 10010;
+int n, m;
+int v;
+int f[N][M];
+
+int main() {
+    cin >> n >> m;
+
+    for (int i = 0; i <= n; i++) f[i][0] = 1; // base case
+
+    for (int i = 1; i <= n; i++) {
+        cin >> v;
+        for (int j = 1; j <= m; j++) {
+            // 从前i-1个物品中选择，装满j这么大的空间，假设方案数是5个
+            // 那么，在前i个物品中选择，装满j这么大的空间，方案数最少也是5个
+            // 如果第i个物品，可以选择，那么可能使得最终的选择方案数增加
+            f[i][j] = f[i - 1][j];
+            // 增加多少呢？前序依赖是：f[i - 1][j - v]
+            if (j >= v) f[i][j] += f[i - 1][j - v];
+        }
+    }
+    // 输出结果
+    printf("%d\n", f[n][m]);
+    return 0;
+}
+```
+
+**一维代码**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 10010;
+
+int n, m;
+int v;
+int f[N]; // 在前i个物品，体积是j的情况下，恰好装满的方案数
+
+int main() {
+    cin >> n >> m;
+
+    // 体积恰好j, f[0]=1, 其余是0
+    f[0] = 1;
+    for (int i = 1; i <= n; i++) {
+        cin >> v;
+        for (int j = m; j >= v; j--)
+            f[j] += f[j - v];
+    }
+    printf("%d\n", f[m]);
+    return 0;
+}
+```
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1010.md b/TangDou/AcWing/LIS/1010.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1010.md
rename to TangDou/AcWing/LIS/1010.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1010_Dilworth.cpp b/TangDou/AcWing/LIS/1010_Dilworth.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1010_Dilworth.cpp
rename to TangDou/AcWing/LIS/1010_Dilworth.cpp
diff --git a/TangDou/AcWing/LIS/1010_TanXin.cpp b/TangDou/AcWing/LIS/1010_TanXin.cpp
new file mode 100644
index 0000000..0dbfd95
--- /dev/null
+++ b/TangDou/AcWing/LIS/1010_TanXin.cpp
@@ -0,0 +1,31 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 1010;
+
+int a[N], f[N]; // 每个导弹的高度，f[i]:以a[i]结尾的最长不升子序列长度
+int q[N], ql;   // 需要ql套拦截系统，每个拦截系统最后一个拦截高度为q[i]
+int n, res;
+
+int main() {
+    while (cin >> a[n]) n++;
+    // 第一问
+    for (int i = 0; i < n; i++) {
+        f[i] = 1;
+        for (int j = 0; j < i; j++)
+            if (a[i] <= a[j]) // 如果前面的比当前的大，说明是不升序列
+                f[i] = max(f[i], f[j] + 1);
+        res = max(res, f[i]);
+    }
+    // 第二问
+    for (int i = 0; i < n; i++) {
+        int p = lower_bound(q, q + ql, a[i]) - q;
+        if (p == ql) // 如果没有找到可以接在后面的导弹拦截系统，那么需要创建一套新的拦截系统
+            q[ql++] = a[i];
+        else
+            q[p] = a[i]; // 否则就直接接到找到的第k套拦截系统的后面,那么，第k套拦截系统的最后一个拦截高度=q[k]=h[i]
+    }
+    // 输出最长不升序列长度,即：最多能拦截的导弹数
+    printf("%d\n%d\n", res, ql);
+    return 0;
+}
\ No newline at end of file
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1012.md b/TangDou/AcWing/LIS/1012.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1012.md
rename to TangDou/AcWing/LIS/1012.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1012_1.cpp b/TangDou/AcWing/LIS/1012_1.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1012_1.cpp
rename to TangDou/AcWing/LIS/1012_1.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1012_2.cpp b/TangDou/AcWing/LIS/1012_2.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1012_2.cpp
rename to TangDou/AcWing/LIS/1012_2.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014.md b/TangDou/AcWing/LIS/1014.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1014.md
rename to TangDou/AcWing/LIS/1014.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014_1.cpp b/TangDou/AcWing/LIS/1014_1.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1014_1.cpp
rename to TangDou/AcWing/LIS/1014_1.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp b/TangDou/AcWing/LIS/1014_2.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp
rename to TangDou/AcWing/LIS/1014_2.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014_3.cpp b/TangDou/AcWing/LIS/1014_3.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1014_3.cpp
rename to TangDou/AcWing/LIS/1014_3.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1016.cpp b/TangDou/AcWing/LIS/1016.cpp
similarity index 87%
rename from TangDou/AcWing_TiGao/T1/LIS/1016.cpp
rename to TangDou/AcWing/LIS/1016.cpp
index 1084f43..5cabba0 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/1016.cpp
+++ b/TangDou/AcWing/LIS/1016.cpp
@@ -5,7 +5,7 @@ using namespace std;
 int n;
 const int N = 100010;
 int a[N];
-int f[N];
+int f[N];//以第i个元素结尾的最大子序列和
 int res;
 
 int main() {
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1016.md b/TangDou/AcWing/LIS/1016.md
similarity index 95%
rename from TangDou/AcWing_TiGao/T1/LIS/1016.md
rename to TangDou/AcWing/LIS/1016.md
index 6499db0..c3a16b3 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/1016.md
+++ b/TangDou/AcWing/LIS/1016.md
@@ -54,7 +54,6 @@ $f[i]$:以第$i$个元素结尾的最大子序列和
 ### 三、$O(n^2)$实现代码
 
 ```cpp {.line-numbers}
-//运行时间：	31 ms
 #include <bits/stdc++.h>
 using namespace std;
 
@@ -62,7 +61,7 @@ using namespace std;
 int n;
 const int N = 100010;
 int a[N];
-int f[N];
+int f[N];//以第i个元素结尾的最大子序列和
 int res;
 
 int main() {
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1017.md b/TangDou/AcWing/LIS/1017.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1017.md
rename to TangDou/AcWing/LIS/1017.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1017_PuSu.cpp b/TangDou/AcWing/LIS/1017_PuSu.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/1017_PuSu.cpp
rename to TangDou/AcWing/LIS/1017_PuSu.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1017_TanXin.cpp b/TangDou/AcWing/LIS/1017_TanXin.cpp
similarity index 61%
rename from TangDou/AcWing_TiGao/T1/LIS/1017_TanXin.cpp
rename to TangDou/AcWing/LIS/1017_TanXin.cpp
index b965294..5fadd0f 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/1017_TanXin.cpp
+++ b/TangDou/AcWing/LIS/1017_TanXin.cpp
@@ -21,32 +21,27 @@ int main() {
         fl = 0, gl = 0;
 
         cin >> n;
-        for (int i = 0; i < n; i++) cin >> a[i];
+        for (int i = 1; i <= n; i++) cin >> a[i];
 
         // 正着求
-        f[0] = a[0];
-        for (int i = 1; i < n; i++) {
+        for (int i = 1; i <= n; i++) {
             if (a[i] > f[fl])
                 f[++fl] = a[i];
             else
-                *lower_bound(f, f + fl, a[i]) = a[i];
+                *lower_bound(f + 1, f + 1 + fl, a[i]) = a[i];
         }
 
-        // 前半程的结果
-        res = fl;
-
         // 反着求
-        g[0] = a[n - 1];
-        for (int i = n - 1; i >= 0; i--) {
+        for (int i = n; i; i--) {
             if (a[i] > g[gl])
                 g[++gl] = a[i];
             else
-                *lower_bound(g, g + gl, a[i]) = a[i];
+                *lower_bound(g + 1, g + 1 + gl, a[i]) = a[i];
         }
         // pk的最大结果
-        res = max(res, gl);
+        res = max(fl, gl);
         // 输出
-        printf("%d\n", res + 1);
+        printf("%d\n", res);
     }
     return 0;
 }
\ No newline at end of file
diff --git a/TangDou/AcWing_TiGao/T1/LIS/187.cpp b/TangDou/AcWing/LIS/187.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/187.cpp
rename to TangDou/AcWing/LIS/187.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/187.md b/TangDou/AcWing/LIS/187.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/187.md
rename to TangDou/AcWing/LIS/187.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/272.md b/TangDou/AcWing/LIS/272.md
similarity index 69%
rename from TangDou/AcWing_TiGao/T1/LIS/272.md
rename to TangDou/AcWing/LIS/272.md
index bea07bf..903fe59 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/272.md
+++ b/TangDou/AcWing/LIS/272.md
@@ -44,9 +44,10 @@ $1≤N≤3000$,序列中的数字均不超过 $2^{31}−1$。
 
 ### 二、前导知识
 
-[$AcWing$ $895$. 最长上升子序列](https://www.cnblogs.com/littlehb/p/15425546.html)
+**[$AcWing$ $895$. 最长上升子序列](https://www.cnblogs.com/littlehb/p/15425546.html)**
 
-**状态表示** $f[i]$表示从第一个数字开始算，以$a[i]$ **<font color='red'>结尾</font>** 的最长的上升序列长度。(以$a[i]$结尾的所有上升序列中属性为**最长**的那一个)
+**状态表示** 
+$f[i]$表示从第一个数字开始算，以$a[i]$ **<font color='red'>结尾</font>** 的最长的上升序列长度。(以$a[i]$结尾的所有上升序列中属性为**最长**的那一个)
 
 **状态计算** 
 
@@ -57,7 +58,7 @@ f[i] = max(f[i], f[j] + 1) & 0 \le j<i \ \& \  a[j]<a[i]
 $$
 
 
-[$AcWing$ $897$. 最长公共子序列](https://www.cnblogs.com/littlehb/p/15429296.html)
+**[$AcWing$ $897$. 最长公共子序列](https://www.cnblogs.com/littlehb/p/15429296.html)**
 
 定义$f[i][j]$是$a[]$以$i$结尾，$b[]$以$j$结尾的 **最长公共子序列长度**，但没有说$a[i]$或者$b[j]$一定要出现在最长公共子序列当中！这个最长公共子序列，可能是$a[]$和$b[]$的一些前序组成的，$a[i],b[j]$也可能没有对结果产生贡献。
 
@@ -68,17 +69,6 @@ f[i][j]=max(f[i-1][j],f[i][j-1]) & a[i] \neq b[j]
 \end{array}\right.
 $
 
-这是两个经典$DP$模型的结合版：
-
-$LIS$ (最长上升子序列，$Longest$ $Increasing$ $Subsequence$)
-
-$LCS$ (最长公共子序列，$Longest$ $Common$ $Subsequence$)
-
-$LCIS$ (最长公共上升子序列，$Longest$ $Common$ $Increasing$ $Subsequence$)
-
-$LCIS$ 也是一个 **相当经典** 的$DP$模型，他的 **状态分析** 是 $LIS$ 与 $LCS$ 的结合，且听我慢慢道来
-
-
 ### 三、本题解法
 闫氏$DP$分析法：(结合了$LCS$与$LIS$的状态表示的方法，可以很直接的发现二者的影子)
 
@@ -91,29 +81,14 @@ $LCIS$ 也是一个 **相当经典** 的$DP$模型，他的 **状态分析** 是
 #### 状态转移
 - $\large a_i \neq b_j$
  考虑 $a$ 数组中前$i-1$个数字， $b$数组中前$j$个数字 ，且当前以 $b[j]$ 结尾的子序列的方案转移过来：
-$$\large f_{i,j}=max(f_{i,j},f_{i−1,j}), a_i \neq b_j$$
+$$\large f[i][j]=max(f[i][j],f[i−1][j])$$
 
 - $\large a_i = b_j$
  考虑 $a$ 数组中前$i-1$个数字， $b$数组中前 $k$ 个数字 ，且当前以 $b[k]$ 结尾的子序列的方案转移过来：
-$$\large f_{i,j}=max(f_{i,j},f_{i−1,k}+1),k∈[0,j−1],a_i=b_j,b_j>b_k$$
+$$\large f[i][j]=max(f[i][j],f[i−1][k]+1),k∈[0,j−1],a_i=b_j,b_j>b_k$$
 
-如果直接按上述思路实现，需要三重循环：
+按上述思路实现，需要三重循环：
 
-```cpp {.line-numbers}
-for (int i = 1; i <= n; i ++ ){
-    for (int j = 1; j <= n; j ++ ){
-        f[i][j] = f[i - 1][j]; //不管a[i]是否等于b[j],f[i][j]一定会从f[i-1][j]继承过来
-        if (a[i] == b[j]){
-            int maxv = 1; //最起码命中了一个a[i]==b[j],LCIS最少是1
-            for (int k = 1; k < j; k ++ )
-                if (a[i] > b[k]) // 不光是公共，还要上升
-                    maxv = max(maxv, f[i - 1][k] + 1);
-            f[i][j] = max(f[i][j], maxv);
-        }
-    }
-}
-``` 
-#### 实现代码$O(N^3)$
 ```cpp {.line-numbers}
 #include <bits/stdc++.h>
 using namespace std;
@@ -122,7 +97,7 @@ const int N = 3010;
 int a[N], b[N];
 int f[N][N];
 int res;
-
+// 通过了 10/13个数据
 // O(n^3)
 int main() {
     int n;
@@ -138,14 +113,14 @@ int main() {
             f[i][j] = f[i - 1][j];
             // ③ 如果恰好 a[i]==b[j],那么就可以发生转移
             if (a[i] == b[j]) {
-                int maxv = 1; // 最起码a[i]==b[j]，有一个数字是一样嘀~
+                int mx = 1; // 最起码a[i]==b[j]，有一个数字是一样嘀~
                 // f[i-1]是肯定的了，问题是b的前驱在哪里？需要枚举1~j-1
                 for (int k = 1; k < j; k++)
-                    if (a[i] > b[k]) // 公共还不成，还需要上升
+                    if (b[j] > b[k]) // 需要上升
                         // 找出公共且最长的
-                        maxv = max(maxv, f[i - 1][k] + 1);
+                        mx = max(mx, f[i - 1][k] + 1);
                 // 更新答案
-                f[i][j] = max(f[i][j], maxv);
+                f[i][j] = max(f[i][j], mx);
             }
         }
     int res = 0;
@@ -158,9 +133,9 @@ int main() {
 
 ###　四、优化
 
-**$Q$:朴素办法超时($10/16$)，如何优化？**
+**$Q$:朴素办法超时($10/13$)，如何优化？**
 
-观察到，对于第二种状态转移：$f_{i,j}=max(f_{i,j},f_{i−1,k}+1) \ k∈[0,j−1],a_i=b_j,b_j>b_k$
+观察到，对于第二种状态转移：$f[i][j]=max(f[i][j],f[i−1][k]+1) \ k∈[0,j−1],a_i=b_j,b_j>b_k$
 
 每次用到的 **状态** 都是第 $i - 1$ 个阶段的
 
@@ -186,11 +161,11 @@ int main() {
     for (int i = 1; i <= n; i++) cin >> b[i];
 
     for (int i = 1; i <= n; i++) {
-        int maxv = 1;
+        int mx = 1;
         for (int j = 1; j <= n; j++) {
             f[i][j] = f[i - 1][j];
-            if (a[i] == b[j]) f[i][j] = max(f[i][j], maxv);
-            if (a[i] > b[j]) maxv = max(maxv, f[i - 1][j] + 1);
+            if (a[i] == b[j]) f[i][j] = mx;
+            if (a[i] > b[j]) mx = max(mx, f[i - 1][j] + 1);
         }
     }
 
@@ -199,6 +174,7 @@ int main() {
 
     return 0;
 }
+
 ```
 
 ### 五、空间优化
diff --git a/TangDou/AcWing_TiGao/T1/LIS/272_1.cpp b/TangDou/AcWing/LIS/272_1.cpp
similarity index 76%
rename from TangDou/AcWing_TiGao/T1/LIS/272_1.cpp
rename to TangDou/AcWing/LIS/272_1.cpp
index 2695705..af9888f 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/272_1.cpp
+++ b/TangDou/AcWing/LIS/272_1.cpp
@@ -5,7 +5,7 @@ const int N = 3010;
 int a[N], b[N];
 int f[N][N];
 int res;
-
+// 通过了 10/13个数据
 // O(n^3)
 int main() {
     int n;
@@ -21,14 +21,13 @@ int main() {
             f[i][j] = f[i - 1][j];
             // ③ 如果恰好 a[i]==b[j],那么就可以发生转移
             if (a[i] == b[j]) {
-                int maxv = 1; // 最起码a[i]==b[j]，有一个数字是一样嘀~
+                int mx = 1; // 最起码a[i]==b[j]，有一个数字是一样嘀~
                 // f[i-1]是肯定的了，问题是b的前驱在哪里？需要枚举1~j-1
                 for (int k = 1; k < j; k++)
-                    if (a[i] > b[k]) // 公共还不成，还需要上升
-                        // 找出公共且最长的
-                        maxv = max(maxv, f[i - 1][k] + 1);
+                    if (a[i] > b[k]) // j可以接在k后面，那么可能的最大值为f[i-1][k]+1
+                        mx = max(mx, f[i - 1][k] + 1);
                 // 更新答案
-                f[i][j] = max(f[i][j], maxv);
+                f[i][j] = max(f[i][j], mx);
             }
         }
     int res = 0;
diff --git a/TangDou/AcWing/LIS/272_2.cpp b/TangDou/AcWing/LIS/272_2.cpp
new file mode 100644
index 0000000..b2b4152
--- /dev/null
+++ b/TangDou/AcWing/LIS/272_2.cpp
@@ -0,0 +1,30 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 3010;
+int a[N], b[N];
+int f[N][N];
+int res;
+
+// O(n^2)
+// f[i][j]—集合：考虑 a 中前 i 个数字，b 中前 j 个数字 ，且当前以 b[j] 结尾的子序列的方案
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= n; i++) cin >> b[i];
+
+    for (int i = 1; i <= n; i++) {
+        int mx = 1; // 如果a[i]==b[j],那么LICS最小是1.如果下面的循环中没有一个if被执行，则mx没使上        
+        for (int j = 1; j <= n; j++) {
+            f[i][j] = f[i - 1][j]; // 先继承过来，现实含义：即使a[i]!=b[j],那么最长长度不会因为i的增加而变化，即f[i][j]=f[i-1][j]
+            if (a[i] == b[j]) f[i][j] = mx;
+            if (a[i] > b[j]) mx = max(mx, f[i - 1][j] + 1);
+        }
+    }
+
+    for (int i = 1; i <= n; i++) res = max(res, f[n][i]);
+    printf("%d\n", res);
+
+    return 0;
+}
diff --git a/TangDou/AcWing_TiGao/T1/LIS/272_YiWei.cpp b/TangDou/AcWing/LIS/272_YiWei.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/272_YiWei.cpp
rename to TangDou/AcWing/LIS/272_YiWei.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/3549.cpp b/TangDou/AcWing/LIS/3549.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/3549.cpp
rename to TangDou/AcWing/LIS/3549.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/3549.eddx b/TangDou/AcWing/LIS/3549.eddx
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/3549.eddx
rename to TangDou/AcWing/LIS/3549.eddx
diff --git a/TangDou/AcWing_TiGao/T1/LIS/3549.md b/TangDou/AcWing/LIS/3549.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/3549.md
rename to TangDou/AcWing/LIS/3549.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/3662.cpp b/TangDou/AcWing/LIS/3662.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/3662.cpp
rename to TangDou/AcWing/LIS/3662.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/3662.md b/TangDou/AcWing/LIS/3662.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/3662.md
rename to TangDou/AcWing/LIS/3662.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/482.md b/TangDou/AcWing/LIS/482.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/482.md
rename to TangDou/AcWing/LIS/482.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/482_1.cpp b/TangDou/AcWing/LIS/482_1.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/482_1.cpp
rename to TangDou/AcWing/LIS/482_1.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/482_2.cpp b/TangDou/AcWing/LIS/482_2.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/482_2.cpp
rename to TangDou/AcWing/LIS/482_2.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/482_3.cpp b/TangDou/AcWing/LIS/482_3.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/482_3.cpp
rename to TangDou/AcWing/LIS/482_3.cpp
diff --git a/TangDou/AcWing/LineDP/895.cpp b/TangDou/AcWing/LIS/895.cpp
similarity index 57%
rename from TangDou/AcWing/LineDP/895.cpp
rename to TangDou/AcWing/LIS/895.cpp
index 63a4b87..79b9b0f 100644
--- a/TangDou/AcWing/LineDP/895.cpp
+++ b/TangDou/AcWing/LIS/895.cpp
@@ -12,9 +12,10 @@ int main() {
     for (int i = 1; i <= n; i++) cin >> a[i];
 
     for (int i = 1; i <= n; i++) {
-        f[i] = 1;
+        f[i] = 1; // 一个以i结尾的序列，最少可以是i自己，长度是1
         for (int j = 1; j < i; j++)
-            if (a[i] > a[j]) f[i] = max(f[i], f[j] + 1);
+            if (a[i] > a[j])                // i可以接在j后面
+                f[i] = max(f[i], f[j] + 1); // 借力于j
         res = max(res, f[i]);
     }
     printf("%d", res);
diff --git a/TangDou/AcWing/LineDP/895.md b/TangDou/AcWing/LIS/895.md
similarity index 100%
rename from TangDou/AcWing/LineDP/895.md
rename to TangDou/AcWing/LIS/895.md
diff --git a/TangDou/AcWing/LineDP/895_WithPrintPath.cpp b/TangDou/AcWing/LIS/895_WithPrintPath.cpp
similarity index 100%
rename from TangDou/AcWing/LineDP/895_WithPrintPath.cpp
rename to TangDou/AcWing/LIS/895_WithPrintPath.cpp
diff --git a/TangDou/AcWing/LineDP/896.cpp b/TangDou/AcWing/LIS/896.cpp
similarity index 100%
rename from TangDou/AcWing/LineDP/896.cpp
rename to TangDou/AcWing/LIS/896.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896.md b/TangDou/AcWing/LIS/896.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896.md
rename to TangDou/AcWing/LIS/896.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_ErFen.cpp b/TangDou/AcWing/LIS/896_ErFen.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_ErFen.cpp
rename to TangDou/AcWing/LIS/896_ErFen.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_ErFen_Inc_Dec.cpp b/TangDou/AcWing/LIS/896_ErFen_Inc_Dec.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_ErFen_Inc_Dec.cpp
rename to TangDou/AcWing/LIS/896_ErFen_Inc_Dec.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_0.cpp b/TangDou/AcWing/LIS/896_FenWickTree_0.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_0.cpp
rename to TangDou/AcWing/LIS/896_FenWickTree_0.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_1.cpp b/TangDou/AcWing/LIS/896_FenWickTree_1.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_1.cpp
rename to TangDou/AcWing/LIS/896_FenWickTree_1.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_2.cpp b/TangDou/AcWing/LIS/896_FenWickTree_2.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_2.cpp
rename to TangDou/AcWing/LIS/896_FenWickTree_2.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_3.cpp b/TangDou/AcWing/LIS/896_FenWickTree_3.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_FenWickTree_3.cpp
rename to TangDou/AcWing/LIS/896_FenWickTree_3.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_WithPrintPath.cpp b/TangDou/AcWing/LIS/896_WithPrintPath.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_WithPrintPath.cpp
rename to TangDou/AcWing/LIS/896_WithPrintPath.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896_使用树状数组优化LIS问题.md b/TangDou/AcWing/LIS/896_使用树状数组优化LIS问题.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/896_使用树状数组优化LIS问题.md
rename to TangDou/AcWing/LIS/896_使用树状数组优化LIS问题.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/897.cpp b/TangDou/AcWing/LIS/897.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/897.cpp
rename to TangDou/AcWing/LIS/897.cpp
diff --git a/TangDou/AcWing_TiGao/T1/LIS/897.md b/TangDou/AcWing/LIS/897.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/LIS/897.md
rename to TangDou/AcWing/LIS/897.md
diff --git a/TangDou/AcWing/LIS/LIS+LCS专题.md b/TangDou/AcWing/LIS/LIS+LCS专题.md
new file mode 100644
index 0000000..217347e
--- /dev/null
+++ b/TangDou/AcWing/LIS/LIS+LCS专题.md
@@ -0,0 +1,651 @@
+## $LIS$+$LCS$专题
+
+#### 一、基础题
+**[$AcWing$ $895$. 最长上升子序列](https://www.acwing.com/problem/content/897/)**
+
+**时间复杂度**
+$O(N^2)$
+
+**状态表示**
+$f[i]$:以$a[i]$结尾的最长上升子序列长度
+
+**状态转移**
+$f[i] = max(f[i], f[j] + 1)(a[i] > a[j])$
+
+**初始值**
+$f[i] = 1$ 
+> **解释**：一个以$i$结尾的序列，最少可以是$i$自己，长度是$1$
+
+**代码**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+
+int n;
+int a[N], f[N];
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+
+    for (int i = 1; i <= n; i++) {
+        f[i] = 1; // 一个以i结尾的序列，最少可以是i自己，长度是1
+        for (int j = 1; j < i; j++)
+            if (a[i] > a[j])                // i可以接在j后面
+                f[i] = max(f[i], f[j] + 1); // 借力于j
+        res = max(res, f[i]);
+    }
+    printf("%d", res);
+    return 0;
+}
+    
+```
+
+
+**[$AcWing$ $896$. 最长上升子序列 II ](https://www.acwing.com/problem/content/898/)**
+
+数据量增大:$N<=100000$
+
+**思想**
+对于同样长度的子串，希望它的末端越小越好，这样后面更易扩展它,使数列更长。
+
+**时间复杂度**
+$O(LogN*N)$
+
+**状态表示**
+$f[i]$表示 **长度为$i$** 的递增子序列中，**末尾元素最小** 的是$f[i]$
+
+**答案**
+$fl$
+
+**特点与总结**
+① $f[]$数组是一个单调上升的数组,这是二分的基础
+② 每个长度都争取替换掉前面记录数组中第一个大于等于自己的数字，以保证长度不变的情况下，数字最小。
+
+**学习方法**
+举栗子
+
+**代码**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 100010;
+
+int n, a[N];
+
+// 数组模拟
+int f[N], fl;
+
+int main() {
+    cin >> n;
+    for (int i = 0; i < n; i++) cin >> a[i];
+
+    // 1、首个元素
+    f[0] = a[0];
+
+    // 2、后续元素开始计算
+    for (int i = 1; i < n; i++) {
+        if (a[i] > f[fl])
+            f[++fl] = a[i];
+        else
+            // 利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
+            *lower_bound(f, f + fl, a[i]) = a[i];
+    }
+    // 栈内元素数量就是答案
+    printf("%d\n", fl + 1);
+    return 0;
+}
+  
+```
+
+
+**[$AcWing$ $897$. 最长公共子序列](https://www.acwing.com/problem/content/899/)**
+
+**状态表示**
+$f[i][j]$:$a[]$以$i$结尾，$b[]$以$j$结尾的 **最长公共子序列长度**
+> **说明**：没有说$a[i]$或者$b[j]$一定要出现在最长公共子序列当中！这个最长公共子序列，可能是$a[]$和$b[]$的一些前序组成的，$a[i],b[j]$也可能没有对结果产生贡献。
+
+* 当$a[i]==b[j]$时，看一下两个字符串的前序，发现在少了$a[i],b[j]$后，转化为子问题$f[i-1][j-1]$,问题转化为$$f[i][j]=f[i-1][j-1]+1$$
+
+* 当$a[i] \neq b[j]$时：
+  * 如果$a[i]$不产生贡献，那么把它干掉$f[i-1][j]$
+  * 如果$b[j]$不产生贡献，那么把它干掉$f[i][j-1]$3
+
+**答案**
+$f[n][m]$
+
+**代码**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+int n, m;
+char a[N], b[N];
+int f[N][N];
+
+int main() {
+    // 递推式出现f[i-1][j-1]，如果i,j从0开始会出现负值，所以下标从1开始
+    cin >> n >> m >> a + 1 >> b + 1;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            if (a[i] == b[j])
+                f[i][j] = f[i - 1][j - 1] + 1;
+            else
+                f[i][j] = max(f[i - 1][j], f[i][j - 1]);
+        }
+    printf("%d", f[n][m]);
+    return 0;
+}
+```
+
+#### 二、进阶题
+**[$AcWing$ $1017$. 怪盗基德的滑翔翼](https://www.acwing.com/problem/content/1019/)**
+
+朴素$O(N^2)$版本
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 110;
+int n;    // 楼房的个数
+int w[N]; // 楼房的高度数组
+int f[N]; // LIS结果数组，DP结果
+
+int main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        // 保持好习惯，多组测试数据，记得每次清空结果数组
+        memset(f, 0, sizeof f);
+        int res = 0;
+
+        cin >> n;
+        for (int i = 1; i <= n; i++) cin >> w[i];
+
+        // 从左到右,求一遍最长上升子序列[朴素O(N^2)版本]
+        for (int i = 1; i <= n; i++) {
+            f[i] = 1;
+            for (int j = 1; j < i; j++)
+                if (w[i] > w[j]) f[i] = max(f[i], f[j] + 1);
+            res = max(res, f[i]);
+        }
+
+        // 反向求解 LIS问题
+        for (int i = n; i >= 1; i--) {
+            f[i] = 1;
+            for (int j = n; j > i; j--)
+                if (w[i] > w[j]) f[i] = max(f[i], f[j] + 1);
+            res = max(res, f[i]);
+        }
+
+        printf("%d\n", res);
+    }
+    return 0;
+}
+```
+优化$O(LogN*N)$版本
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 110;
+int n;    // 楼房的个数
+int a[N]; // 楼房的高度数组
+
+// 数组模拟栈
+int f[N], fl;
+int g[N], gl;
+int res;
+
+int main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        // 保持好习惯，多组测试数据，记得每次清空结果数组
+        memset(f, 0, sizeof f);
+        memset(g, 0, sizeof g);
+        fl = 0, gl = 0;
+
+        cin >> n;
+        for (int i = 1; i <= n; i++) cin >> a[i];
+
+        // 正着求
+        for (int i = 1; i <= n; i++) {
+            if (a[i] > f[fl])
+                f[++fl] = a[i];
+            else
+                *lower_bound(f + 1, f + 1 + fl, a[i]) = a[i];
+        }
+
+        // 反着求
+        for (int i = n; i; i--) {
+            if (a[i] > g[gl])
+                g[++gl] = a[i];
+            else
+                *lower_bound(g + 1, g + 1 + gl, a[i]) = a[i];
+        }
+        // pk的最大结果
+        res = max(fl, gl);
+        // 输出
+        printf("%d\n", res);
+    }
+    return 0;
+}
+```
+
+**[$AcWing$ $1014$. 登山](https://www.acwing.com/problem/content/1016/)**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 1010;
+int n;    // 山的个数
+int a[N]; // 山的高度数组
+int f[N]; // 最长上升子序列
+int g[N]; // 最长下降子序列
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+
+    // 正向
+    for (int i = 1; i <= n; i++) {
+        f[i] = 1;
+        for (int j = 1; j < i; j++)
+            if (a[i] > a[j]) f[i] = max(f[i], f[j] + 1);
+    }
+    // 反向
+    for (int i = n; i >= 1; i--) {
+        g[i] = 1;
+        for (int j = n; j > i; j--)
+            if (a[i] > a[j]) g[i] = max(g[i], g[j] + 1);
+    }
+
+    // 每个点，都可能是两者相加的最大位置处，所以，需要枚举每个点，每个点都有资格参评最优点
+    // 因为最终的那个中间点，左边计算了一次，右边又计算了一次，需要减1去重复
+    for (int i = 1; i <= n; i++) res = max(res, f[i] + g[i] - 1);
+    // 输出
+    printf("%d\n", res);
+    return 0;
+}
+```
+
+**[$AcWing$ $482$. 合唱队形](https://www.acwing.com/problem/content/484/)**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1010;
+int n, a[N];
+int f[N], fl, p1[N];
+int g[N], gl, p2[N];
+
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+
+    // 正向
+    f[++fl] = a[1];
+    p1[1] = 1;
+
+    for (int i = 2; i <= n; i++)
+        if (a[i] > f[fl]) {
+            f[++fl] = a[i];
+            p1[i] = fl;
+        } else {
+            int t = lower_bound(f + 1, f + fl + 1, a[i]) - f;
+            f[t] = a[i];
+            p1[i] = t;
+        }
+
+    // 反向
+    g[++gl] = a[n];
+    p2[n] = 1;
+
+    for (int i = n - 1; i >= 1; i--)
+        if (a[i] > g[gl]) {
+            g[++gl] = a[i];
+            p2[i] = gl;
+        } else {
+            int t = lower_bound(g + 1, g + gl + 1, a[i]) - g;
+            g[t] = a[i];
+            p2[i] = t;
+        }
+
+    for (int i = 1; i <= n; i++) res = max(res, p2[i] + p1[i] - 1);
+
+    printf("%d\n", n - res);
+    return 0;
+}
+```
+**[$AcWing$ $1012$. 友好城市](https://www.acwing.com/problem/content/1014/)**
+① 数对 
+② 左端点排序
+③ 对于右端点组成的数组，求最长上升子序列长度
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef pair<int, int> PII;
+#define x first
+#define y second
+const int N = 5010;
+PII a[N]; // 南岸和北岸的一对友好城市的坐标
+int f[N]; // 记录以f[i]为结尾的一对友好城市时的，不产生交叉的最多城市对数
+int n;    // n组友好城市
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
+    sort(a + 1, a + 1 + n); // PII默认按第一维度first进行排序
+
+    // LIS
+    for (int i = 1; i <= n; i++) {
+        f[i] = 1;
+        for (int j = 1; j < i; j++)
+            if (a[i].y > a[j].y)
+                f[i] = max(f[i], f[j] + 1);
+        res = max(res, f[i]);
+    }
+    printf("%d\n", res);
+    return 0;
+}
+```
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef pair<int, int> PII;
+#define x first
+#define y second
+
+const int N = 5010;
+PII a[N];     // 南岸和北岸的一对友好城市的坐标
+int f[N], fl; // 数组模拟栈
+int n;        // n组友好城市
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
+    sort(a + 1, a + 1 + n);
+
+    f[++fl] = a[1].y;
+    for (int i = 2; i <= n; i++) {
+        if (a[i].y > f[fl])
+            f[++fl] = a[i].y;
+        else
+            *lower_bound(f + 1, f + 1 + fl, a[i].y) = a[i].y;
+    }
+
+    printf("%d\n", fl);
+    return 0;
+}
+```
+
+**[$AcWing$ $1016$. 最大上升子序列和](https://www.acwing.com/problem/content/1012/)**
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+// 最大上升子序列和
+int n;
+const int N = 100010;
+int a[N];
+int f[N];//以第i个元素结尾的最大子序列和
+int res;
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+
+    for (int i = 1; i <= n; i++) {
+        f[i] = a[i]; // 最大上升子序列（个数）这里是1,此处是a[i]
+        for (int j = 1; j < i; j++)
+            // 最大上升子序列（个数）这里是加1,此处是+a[i]
+            if (a[i] > a[j]) f[i] = max(f[i], f[j] + a[i]);
+        res = max(res, f[i]);
+    }
+    printf("%d ", res);
+    return 0;
+}
+```
+**魔改 最长上升子序列和**
+
+**[$AcWing$ $1010$. 拦截导弹](https://www.acwing.com/problem/content/1012/)**
+
+> **题目要求**：但是这种导弹拦截系统有一个缺陷：虽然它的第一发炮弹能够到达任意的高度，但是以后每一发炮弹都不能高于前一发的高度。
+
+**数组含义**
+$q[],ql$:已经创建了$ql$套导弹系统,$q[i]$中记录的是第$i$套导弹系统的最大拦截高度
+
+**举栗子,总结规律**
+$q[0]:5$
+$q[1]:6$
+> **举的这个栗子对吗？**
+> 我们思考一下：原来有一套导弹拦截系统，上次的高度是$5$,现在第二颗导弹的高度是$6$，根据题目要求，它不能高于前一发的高度，所以只能再开一个导弹拦截系统，所以，第二个是$6$是正确的。
+
+现在来了一个高度=$4$,根据上面的原则，可以接在$5$后面，也可以接在$6$后面，我们为了不浪费$6$,就选择了接在$5$后面，放到$q[0]$里，修改$q[0]=4$
+$q[0]:4$
+$q[1]:6$
+原来的数组是单调上升的，修改后值也是在上下夹着中间的，也依然是单调上升的。
+
+再比如：
+$q[0]:4$
+$q[1]:6$
+$q[2]:7$
+现在来了一个$a[i]=5$,我们肯定要修改$q[1]=5$,变为：
+
+$q[0]:4$
+$q[1]:5$
+$q[2]:7$
+原来的数组是 **单调上升** 的，修改后也依然是 **单调上升** 的。
+
+既然有这个规律，那么就可以使用二分快速查找大于等于$a[i]$的位置$p$了：
+        
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 1010;
+
+int a[N], f[N]; // 每个导弹的高度，f[i]:以a[i]结尾的最长不升子序列长度
+int q[N], ql;   // 需要ql套拦截系统，每个拦截系统最后一个拦截高度为q[i]
+int n, res;
+
+int main() {
+    while (cin >> a[n]) n++;
+    // 第一问
+    for (int i = 0; i < n; i++) {
+        f[i] = 1;
+        for (int j = 0; j < i; j++)
+            if (a[i] <= a[j]) // 如果前面的比当前的大，说明是不升序列
+                f[i] = max(f[i], f[j] + 1);
+        res = max(res, f[i]);
+    }
+    // 第二问
+    for (int i = 0; i < n; i++) {
+        int p = lower_bound(q, q + ql, a[i]) - q;
+        if (p == ql) // 如果没有找到可以接在后面的导弹拦截系统，那么需要创建一套新的拦截系统
+            q[ql++] = a[i];
+        else
+            q[p] = a[i]; // 否则就直接接到找到的第k套拦截系统的后面,那么，第k套拦截系统的最后一个拦截高度=q[k]=h[i]
+    }
+    // 输出最长不升序列长度,即：最多能拦截的导弹数
+    printf("%d\n%d\n", res, ql);
+    return 0;
+}  
+```
+
+**[$AcWing$ $187$. 导弹防御系统](https://www.acwing.com/problem/content/189/)**
+
+导弹拦截系统的拦截高度即可以 严格单调上升，又可以严格单调下降，此时，用$LIS$模型是无法解决问题的，考虑到$n<50$，可以用$dfs+$剪枝
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 55;
+const int INF = 0x3f3f3f3f;
+int n;
+int a[N];
+int up[N], down[N];
+int res;
+
+// 本题关键字：贪心+爆搜
+
+/*
+ 功能：暴搜所有可能，配合剪枝，找出最少的拦截系统数量
+ u:  第几个导弹
+ ul: 上升拦截系统的数量，配合up  数组使用
+ dl: 下降拦截系统的数量，配合down数组使用
+ */
+void dfs(int u, int ul, int dl) {
+    if (ul + dl >= res) return;  // 伟大的剪枝，不剪枝会TLE~,中途发现已经大于等于res的情况，就返回
+    if (u == n) {                // 走完全程,收集答案
+        res = min(res, ul + dl); // 因为上面的剪枝，把ul+dl>=res的全干掉了，能到这里的，都是<res的，都可以用来更新答案
+        return;
+    }
+
+    // 放入上升序列
+    int k = 0;
+    // 如果把当前导弹使用上升序列的拦截系统进行拦截，那个选择哪个系统呢？
+    for (k = 0; k < ul; k++)
+        if (up[k] <= a[u]) break; // up[0],up[1],up[2],... 这样套拦截系统，其数值来讲，是递减的，因为是因为不能再拦截高度更低的才创建了一套新的拦截系统
+    // 找出放到哪个拦截系统上去，结果为k
+
+    int t = up[k]; // 尝试在第k套系统进行拦截第u个导弹,那么，第u个导弹的高度就是第k套系统的新高度
+    up[k] = a[u];  // 问题是，我们也不一定非得让第u个导弹使用上升序列拦截系统，也可能使用下降序列拦截系统，所以需要记录当前值，回溯后，尝试下降序列拦截
+
+    if (k < ul)                 // 如果当前这些套拦截系统中可以找到某一套进行拦截
+        dfs(u + 1, ul, dl);     // 接到了某个队伍后面去了，修改队伍的最后最大值即可，队伍数量没有长大
+    else                        // 如果当前这些套拦截系统中无法找到某一套进行拦截
+        dfs(u + 1, ul + 1, dl); // 没有找到任何一个符合最后一个高度小于a[u]的队伍，只能多开一队，up数组长度长大1
+
+    up[k] = t; // 回溯
+
+    // ----------------------------------------------------------------------
+
+    // 放入下降序列
+    k = 0;
+    for (k = 0; k < dl; k++)
+        if (down[k] >= a[u]) break;
+
+    t = down[k]; // 保留现场
+    down[k] = a[u];
+    if (k < dl)
+        dfs(u + 1, ul, dl);
+    else
+        dfs(u + 1, ul, dl + 1);
+    down[k] = t; // 回溯
+}
+
+int main() {
+    while (cin >> n, n) { // 多套数据，输入n=0时停止
+        for (int i = 0; i < n; i++) cin >> a[i];
+        res = INF;    // 防御系统的最少数量
+        dfs(0, 0, 0); // 开始深搜，更新res的值
+        printf("%d\n", res);
+    }
+    return 0;
+}
+
+```
+
+
+**[$AcWing$ $272$. 最长公共上升子序列](https://www.acwing.com/problem/content/274/)**
+
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 3010;
+int a[N], b[N];
+int f[N][N];
+int res;
+// 通过了 10/13个数据
+// O(n^3)
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= n; i++) cin >> b[i];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= n; j++) {
+            // ① 二维DP打表的一般套路，都是可以直接从上一行继承的
+            // ② 从题意出发，就是a中前i个数字，b中前j个数字，且以b[j]结尾的子序列中长度最大的
+            // 那么，a中多整出一个数字来，最起码也是f[i-1][j]的值，不能更小
+            f[i][j] = f[i - 1][j];
+            // ③ 如果恰好 a[i]==b[j],那么就可以发生转移
+            if (a[i] == b[j]) {
+                int mx = 1; // 最起码a[i]==b[j]，有一个数字是一样嘀~
+                // f[i-1]是肯定的了，问题是b的前驱在哪里？需要枚举1~j-1
+                for (int k = 1; k < j; k++)
+                    if (a[i] > b[k]) // j可以接在k后面，那么可能的最大值为f[i-1][k]+1
+                        mx = max(mx, f[i - 1][k] + 1);
+                // 更新答案
+                f[i][j] = max(f[i][j], mx);
+            }
+        }
+    int res = 0;
+    // a数组肯定是火力全开到n就行，b数组中的位置就需要枚举了
+    for (int i = 1; i <= n; i++) res = max(res, f[n][i]);
+    printf("%d\n", res);
+    return 0;
+}
+```
+前面之所以把$b[j]$换成$a[i]$,是因为方便现在考虑。
+
+由于在枚举计算$f[i][j]$时，却用到的是和内层循环（即$b[j]$）无关的，只和外层循环（即$a[i]$）有关，而我们在计算$f[i][j]$，每次都要用到前$j−1$个数中小于$a[i]$的$f[i-1][k]$最大值，所以我们可以用一个变量$mx$不断更新它来保存前$j−1$的答案,就可以避免重复计算，从而降低时间复杂度。
+
+$O(N^2)$
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 3010;
+int a[N], b[N];
+int f[N][N];
+int res;
+
+// O(n^2)
+// f[i][j]—集合：考虑 a 中前 i 个数字，b 中前 j 个数字 ，且当前以 b[j] 结尾的子序列的方案
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 1; i <= n; i++) cin >> b[i];
+
+    for (int i = 1; i <= n; i++) {
+        int mx = 1; // 如果a[i]==b[j],那么LICS最小是1.如果下面的循环中没有一个if被执行，则mx没使上        
+        for (int j = 1; j <= n; j++) {
+            f[i][j] = f[i - 1][j]; // 先继承过来，现实含义：即使a[i]!=b[j],那么最长长度不会因为i的增加而变化，即f[i][j]=f[i-1][j]
+            if (a[i] == b[j]) f[i][j] = mx;
+            if (a[i] > b[j]) mx = max(mx, f[i - 1][j] + 1);
+        }
+    }
+
+    for (int i = 1; i <= n; i++) res = max(res, f[n][i]);
+    printf("%d\n", res);
+
+    return 0;
+}
+```
+[最长上升子序列 ($LIS$) 详解+例题模板 (全)](https://blog.csdn.net/lxt_Lucia/article/details/81206439)
diff --git a/TangDou/AcWing/LineDP/896.md b/TangDou/AcWing/LineDP/896.md
deleted file mode 100644
index 41be38b..0000000
--- a/TangDou/AcWing/LineDP/896.md
+++ /dev/null
@@ -1,210 +0,0 @@
-##[$AcWing$ $896$. 最长上升子序列 II](https://www.acwing.com/problem/content/description/898/)
-
-### 一、题目描述
-
-给定一个长度为 $N$ 的数列，**求数值严格单调递增的子序列的长度最长** 是多少。
-
-**输入格式**
-第一行包含整数 $N$。
-
-第二行包含 $N$ 个整数，表示完整序列。
-
-**输出格式**
-输出一个整数，表示最大长度。
-
-**数据范围**
-$1≤N≤100000$，
-$−10^9≤数列中的数≤10^9$
-
-**输入样例：**
-```cpp {.line-numbers}
-7
-3 1 2 1 8 5 6
-```
-
-**输出样例**：
-```cpp {.line-numbers}
-4
-```
-
-<!-- 让表格居中显示的风格 -->
-<style>
-.center 
-{
-  width: auto;
-  display: table;
-  margin-left: auto;
-  margin-right: auto;
-}
-</style>
-
-### 二、贪心+二分优化
-
-#### 1、与朴素版本的区别
-
-> **朴素版本状态表示**：
-> - 集合:$f[i]$表示从第一个数字开始算，以 $a[i]$<font color='red'><b>   结尾 </b></font>的最长的上升序列长度。
-> - 属性：$max$
- 
-
-前一版本：$N≤1000$，本题要求：$N≤100000$
-
-$N$的数据范围大了$100$倍,前一版本动态规划代码的时间复杂度是$O(N^2)$,$1000^2=1000000$，是$1e6$，是可以$1$秒过的，但如果是$100000^2=10000000000$,是$1e10$，超时，需要要优化~
-
-
-
-#### 2、贪心+二分算法
-
-><font color='red' size=4><b>核心思想:
-对于同样长度的子串，希望它的末端越小越好，这样后面有更多机会拓展它,才有可能使得数列更长。</b></font>
-
-> **状态表示**：
-> - 集合：$f[i]$表示长度为$i$的递增子序列中，末尾元素最小的是$f[i]$
-> - 属性：$min$
-
-**算法步骤**
-* 扫描每个原序列中的数字:
-  * 如果$f$中的最后一个数字$f[idx]$小于当前数字$a[i]$，那么就在$f$的最后面增加$a[i]$
- 
-  * 如果$a[i]$小于$f[idx]$,在$f$中查找并替换第一个大于等于它元素
-  
-**举栗子模拟**
-<div class="center">
-
-| $arr$ | $3$ | $1$ | $2$ | $1$ | $8$ | $5$ | $6$ |
-| ----- | --- | --- | --- | --- | --- | --- | --- |
-
-</div>
-
-
-开始时$f[]$为空，数字$3$进入序列 
-
-<div class="center">
-
-| $arr$ | <font color='red'>$3$</font> | $1$ | $2$ | $1$ | $8$ | $5$ | $6$ |
-| ----- | ---------------------------- | --- | --- | --- | --- | --- | --- |
-
-| $f$ | <font color='red' size=3><b>$3$</b></font> |
-| --- | ------------------------------------------ |
-
-</div>
-
-
-$1$ 比 $3$ 小, $3$出序列 ，$1$入序列 
-
-<div class="center">
-
-| $arr$ | $3$ | <font color='red'> $1$ </font> | $2$ | $1$ | $8$ | $5$ | $6$ |
-| ----- | --- | ------------------------------ | --- | --- | --- | --- | --- |
-
-
-| $f$ | <font color='red' size=4><b>$1$</b></font> |
-| --- | ------------------------------------------ |
-
-</div>
-
-
-$2$ 比 $1$ 大,$2$入序列 
-
-<div class="center">
-
-| $arr$ | $3$ | $1$ | <font color='red'>$2$ </font> | $1$ | $8$ | $5$ | $6$ |
-| ----- | --- | --- | ----------------------------- | --- | --- | --- | --- |
-
-
-| $f$ | $1$ | <font color='red' size=4><b>$2$</b></font> |
-| --- | --- | ------------------------------------------ |
-
-</div>
-
-
-$1$ 比 $2$ 小,在$f$中找到第一个大于等于$1$的位置，并替换掉原来的数字
-
-<div class="center">
-
-| $arr$ | $3$ | $1$ | $2$ | <font color='red'>$1$</font> | $8$ | $5$ | $6$ |
-| ----- | --- | --- | --- | ---------------------------- | --- | --- | --- |
-
-
-| $f$ | <font color='red' size=4><b>$1$</b></font> | $2$ |
-| --- | ------------------------------------------ | --- |
-</div>
-
-
-$8$ 比 $2$ 大
-
-<div class="center">
-
-| $arr$ | $3$ | $1$ | $2$ | $1$ | <font color='red'>$8$</font> | $5$ | $6$ |
-| ----- | --- | --- | --- | --- | ---------------------------- | --- | --- |
-
-
-| $f$ | $1$ | $2$ | <font color='red' size=4><b>$8$</b></font> |
-| --- | --- | --- | ------------------------------------------ |
-</div>
-
-
-$5$ 比 $8$ 小,在$f$中找到第一个大于等于$5$的数字，并替换掉原来的数字
-
-<div class="center">
-
-| $arr$ | $3$ | $1$ | $2$ | $1$ | $8$ | <font color='red'>$5$</font> | $6$ |
-| ----- | --- | --- | --- | --- | --- | ---------------------------- | --- |
-
-
-| $f$ | $1$ | $2$ | <font color='red' size=4><b>$5$</b></font> |
-| --- | --- | --- | ------------------------------------------ |
-</div>
-
-
-$6$ 比 $5$ 大
-
-<div class="center">
-
-| $arr$ | $3$ | $1$ | $2$ | $1$ | $8$ | $5$ | <font color='red' size=4><b>$6$ </b></font> |
-| ----- | --- | --- | --- | --- | --- | --- | ------------------------------------------- |
-
-
-| $f$ | $1$ | $2$ | $5$ | <font color='red' size=4><b>$6$</b></font> |
-| --- | --- | --- | --- | ------------------------------------------ |
-</div>
-
-
-<font color='blue' size=6><b><center>$f$ 的长度$idx$就是最长递增子序列的长度</center></b></font>
-
-
-#### 3、时间复杂度
-$$\large O(N*logN)$$
-
-### 四、实现代码
-```cpp {.line-numbers}
-#include <bits/stdc++.h>
-
-using namespace std;
-const int N = 100010;
-
-int n, a[N];
-
-// 数组模拟栈
-int f[N], fl;
-
-int main() {
-    cin >> n;
-    for (int i = 0; i < n; i++) cin >> a[i];
-
-    // 1、首个元素入栈
-    f[0] = a[0];
-
-    // 2、后续元素开始计算
-    for (int i = 1; i < n; i++) {
-        if (a[i] > f[fl])
-            f[++fl] = a[i];
-        else
-            // 利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
-            *lower_bound(f, f + fl, a[i]) = a[i];
-    }
-    // 栈内元素数量就是答案
-    printf("%d\n", fl + 1);
-    return 0;
-}
-```
\ No newline at end of file
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1015.md b/TangDou/AcWing/NumberTriangles/1015.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1015.md
rename to TangDou/AcWing/NumberTriangles/1015.md
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1015_Dfs.cpp b/TangDou/AcWing/NumberTriangles/1015_Dfs.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1015_Dfs.cpp
rename to TangDou/AcWing/NumberTriangles/1015_Dfs.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1015_Dfs_JiYIHua.cpp b/TangDou/AcWing/NumberTriangles/1015_Dfs_JiYIHua.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1015_Dfs_JiYIHua.cpp
rename to TangDou/AcWing/NumberTriangles/1015_Dfs_JiYIHua.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1015_ErWei.cpp b/TangDou/AcWing/NumberTriangles/1015_ErWei.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1015_ErWei.cpp
rename to TangDou/AcWing/NumberTriangles/1015_ErWei.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1015_YiWei.cpp b/TangDou/AcWing/NumberTriangles/1015_YiWei.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1015_YiWei.cpp
rename to TangDou/AcWing/NumberTriangles/1015_YiWei.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1018.md b/TangDou/AcWing/NumberTriangles/1018.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1018.md
rename to TangDou/AcWing/NumberTriangles/1018.md
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1018_ErWei.cpp b/TangDou/AcWing/NumberTriangles/1018_ErWei.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1018_ErWei.cpp
rename to TangDou/AcWing/NumberTriangles/1018_ErWei.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1018_YiWei.cpp b/TangDou/AcWing/NumberTriangles/1018_YiWei.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1018_YiWei.cpp
rename to TangDou/AcWing/NumberTriangles/1018_YiWei.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1027.md b/TangDou/AcWing/NumberTriangles/1027.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1027.md
rename to TangDou/AcWing/NumberTriangles/1027.md
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1027_0.cpp b/TangDou/AcWing/NumberTriangles/1027_0.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1027_0.cpp
rename to TangDou/AcWing/NumberTriangles/1027_0.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1027_1.cpp b/TangDou/AcWing/NumberTriangles/1027_1.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1027_1.cpp
rename to TangDou/AcWing/NumberTriangles/1027_1.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/1027_2.cpp b/TangDou/AcWing/NumberTriangles/1027_2.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/1027_2.cpp
rename to TangDou/AcWing/NumberTriangles/1027_2.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/275.cpp b/TangDou/AcWing/NumberTriangles/275.cpp
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/275.cpp
rename to TangDou/AcWing/NumberTriangles/275.cpp
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/275.md b/TangDou/AcWing/NumberTriangles/275.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/275.md
rename to TangDou/AcWing/NumberTriangles/275.md
diff --git a/TangDou/AcWing_TiGao/T1/NumberTriangles/数字三角形专题.md b/TangDou/AcWing/NumberTriangles/数字三角形专题.md
similarity index 100%
rename from TangDou/AcWing_TiGao/T1/NumberTriangles/数字三角形专题.md
rename to TangDou/AcWing/NumberTriangles/数字三角形专题.md
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1010_TanXin.cpp b/TangDou/AcWing_TiGao/T1/LIS/1010_TanXin.cpp
deleted file mode 100644
index c79570b..0000000
--- a/TangDou/AcWing_TiGao/T1/LIS/1010_TanXin.cpp
+++ /dev/null
@@ -1,57 +0,0 @@
-#include <bits/stdc++.h>
-using namespace std;
-
-const int N = 1010;
-
-int a[N], f[N]; // 每个导弹的高度，f[i]:以a[i]结尾的最长不升子序列长度
-int q[N], ql;   // 需要ql套拦截系统，每个拦截系统最后一个拦截高度为q[i]
-int n, res;
-
-int main() {
-    while (cin >> a[n]) n++;
-    // 第一问
-    for (int i = 0; i < n; i++) {
-        f[i] = 1;
-        for (int j = 0; j < i; j++)
-            if (a[i] <= a[j]) // 如果前面的比当前的大，说明是不升序列
-                f[i] = max(f[i], f[j] + 1);
-        res = max(res, f[i]);
-    }
-    // 第二问
-    for (int i = 0; i < n; i++) {
-        /* 1、现有已经创建了ql套导弹系统
-           2、算法思路：
-           (1) 在已有的所有导弹拦截系统中找到大于当前高度，并且最小的那个，把当前导弹用这套拦截系统进行拦截
-           (2) 如果找不到这样的拦截系统，说明当前所有拦截系统的最小拦截高度都小于当前导弹，都无法利用，必须新开一个拦截系统。
-           (3) 也就是说，只有前面拦截系统的最后一个拦截高度，都小于当前高度时，才会创建新的，所以，后面的拦截系统，它的尾巴值应该是最大的
-           (4) 比如：
-           q[0]:5
-           q[1]:6
-           现在来了一个高度=4,根据上面的原则，应该放到q[0]里，修改q[0]=4
-           q[0]:4
-           q[1]:6
-           原来的数组是单调上升的，修改后也依然是单调上升的。
-
-           再比如：
-           q[0]:4
-           q[1]:6
-           q[2]:7
-           现在来了一个a[i]=5,我们肯定要修改q[1]=5,变为：
-
-           q[0]:4
-           q[1]:5
-           q[2]:7
-           原来的数组是单调上升的，修改后也依然是单调上升的。
-
-           既然有这个规律，那么就可以使用二分快速查找大于等于a[i]的位置p了：
-        */
-        int p = lower_bound(q, q + ql, a[i]) - q;
-        if (p == ql) // 如果没有找到可以接在后面的导弹拦截系统，那么需要创建一套新的拦截系统
-            q[ql++] = a[i];
-        else
-            q[p] = a[i]; // 否则就直接接到找到的第k套拦截系统的后面,那么，第k套拦截系统的最后一个拦截高度=q[k]=h[i]
-    }
-    // 输出最长不升序列长度,即：最多能拦截的导弹数
-    printf("%d\n%d\n", res, ql);
-    return 0;
-}
\ No newline at end of file
diff --git a/TangDou/AcWing_TiGao/T1/LIS/272_2.cpp b/TangDou/AcWing_TiGao/T1/LIS/272_2.cpp
deleted file mode 100644
index 88309d5..0000000
--- a/TangDou/AcWing_TiGao/T1/LIS/272_2.cpp
+++ /dev/null
@@ -1,29 +0,0 @@
-#include <bits/stdc++.h>
-
-using namespace std;
-const int N = 3010;
-int a[N], b[N];
-int f[N][N];
-int res;
-
-// O(n^2)
-int main() {
-    int n;
-    cin >> n;
-    for (int i = 1; i <= n; i++) cin >> a[i];
-    for (int i = 1; i <= n; i++) cin >> b[i];
-
-    for (int i = 1; i <= n; i++) {
-        int maxv = 1;
-        for (int j = 1; j <= n; j++) {
-            f[i][j] = f[i - 1][j];
-            if (a[i] == b[j]) f[i][j] = maxv;
-            if (a[i] > b[j]) maxv = max(maxv, f[i - 1][j] + 1);
-        }
-    }
-
-    for (int i = 1; i <= n; i++) res = max(res, f[n][i]);
-    printf("%d\n", res);
-
-    return 0;
-}
diff --git a/TangDou/AcWing_TiGao/T1/LIS/896.cpp b/TangDou/AcWing_TiGao/T1/LIS/896.cpp
deleted file mode 100644
index c892b3e..0000000
--- a/TangDou/AcWing_TiGao/T1/LIS/896.cpp
+++ /dev/null
@@ -1,29 +0,0 @@
-#include <bits/stdc++.h>
-
-using namespace std;
-const int N = 100010;
-
-int n, a[N];
-
-// 数组模拟栈
-int f[N], fl;
-
-int main() {
-    cin >> n;
-    for (int i = 0; i < n; i++) cin >> a[i];
-
-    // 1、首个元素入栈
-    f[0] = a[0];
-
-    // 2、后续元素开始计算
-    for (int i = 1; i < n; i++) {
-        if (a[i] > f[fl])
-            f[++fl] = a[i];
-        else
-            // 利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
-            *lower_bound(f, f + fl, a[i]) = a[i];
-    }
-    // 栈内元素数量就是答案
-    printf("%d\n", fl + 1);
-    return 0;
-}
\ No newline at end of file
diff --git a/TangDou/AcWing_TiGao/T1/LIS/LIS+LCS专题.md b/TangDou/AcWing_TiGao/T1/LIS/LIS+LCS专题.md
deleted file mode 100644
index 331b61f..0000000
--- a/TangDou/AcWing_TiGao/T1/LIS/LIS+LCS专题.md
+++ /dev/null
@@ -1,103 +0,0 @@
-## $LIS$+$LCS$专题
-
-#### 基础题
-**[$AcWing$ $895$. 最长上升子序列](https://www.acwing.com/problem/content/897/)**
-
-$O(N^2)$算法
-
-**状态表示**
-$f[i]$:以$a[i]$结尾的最长上升子序列长度
-
-**状态转移**
-```cpp {.line-numbers}
-    for (int i = 1; i <= n; i++) {
-        f[i] = 1; //只有a[i]一个数
-        for (int j = 1; j < i; j++)
-            if (a[j] < a[i]) f[i] = max(f[i], f[j] + 1);
-    }
-```
-**答案**
-```cpp {.line-numbers}
-for (int i = 1; i <= n; i++) res = max(res, f[i]);
-```
-
-
-**[$AcWing$ $896$. 最长上升子序列 II ](https://www.acwing.com/problem/content/898/)**
-
-数据量增大:$N<=100000$
-$O(LogN*N)$算法
-
-```cpp {.line-numbers}
- // 1、首个元素入栈
-f[0] = a[0];
-
-// 2、后续元素开始计算
-for (int i = 1; i < n; i++) {
-    if (a[i] > f[fl])
-        f[++fl] = a[i];
-    else
-        //利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
-        *lower_bound(f, f + fl, a[i]) = a[i];
-}    
-```
-
-也可以这样写
-```cpp {.line-numbers}
-memset(f,-0x3f,sizeof f);
-   
-// 2、后续元素开始计算
-for (int i = 1; i <= n; i++) {
-    if (a[i] > f[fl])
-        f[++fl] = a[i];
-    else
-        //利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
-        *lower_bound(f + 1, f + 1 + fl, a[i]) = a[i];
-}
-```
-
-很明显，第一种写法性能更高，第二种写法有点暴力。
-
-**状态表示**
-$f[i]$表示长度为$i$的递增子序列中，末尾元素最小的是$f[i]$
-
-**答案**
-$fl$
-
-**状态转移**
-① $f[]$数组是一个单独上升的数组,这是可以二分的基础
-② 每个长度都争取替换掉前面记录数组中第一个大于等于自己的数字，以保证长度不变的情况下，数字最小，因为只有最小才能让后面的其它数字更容易接上去，机会才能更多。
-
-**[$AcWing$ $897$. 最长公共子序列](https://www.acwing.com/problem/content/899/)**
-
-**状态表示**
-定义$f[i][j]$是$a[]$以$i$结尾，$b[]$以$j$结尾的**最长公共子序列长度**
-> **说明**：没有说$a[i]$或者$b[j]$一定要出现在最长公共子序列当中！这个最长公共子序列，可能是$a[]$和$b[]$的一些前序组成的，$a[i],b[j]$也可能没有对结果产生贡献。
-
-* 当$a[i]==b[j]$时，看一下两个字符串的前序，发现在少了$a[i],b[j]$后，转化为子问题$f[i-1][j-1]$,问题转化为$$f[i][j]=f[i-1][j-1]+1$$
-
-* 当$a[i] \neq b[j]$时：
-  * 如果$a[i]$不产生贡献，那么把它干掉$f[i-1][j]$
-  * 如果$b[j]$不产生贡献，那么把它干掉$f[i][j-1]$
-```cpp {.line-numbers}
-for (int i = 1; i <= n; i++)
-        for (int j = 1; j <= m; j++) {
-            if (a[i] == b[j])
-                f[i][j] = f[i - 1][j - 1] + 1;
-            else
-                f[i][j] = max(f[i - 1][j], f[i][j - 1]);
-        }
-```
-**答案**
-$f[n][m]$
-
-#### 进阶题
-AcWing 1017. 怪盗基德的滑翔翼
-AcWing 1014. 登山
-AcWing 482. 合唱队形
-AcWing 1012. 友好城市
-AcWing 1016. 最大上升子序列和
-AcWing 1010. 拦截导弹
-AcWing 187. 导弹防御系统
-AcWing 272. 最长公共上升子序列
-
-[最长上升子序列 ($LIS$) 详解+例题模板 (全)](https://blog.csdn.net/lxt_Lucia/article/details/81206439)
diff --git a/TangDou/LanQiaoBei/2023年12月STEMA C++组中高级真题.pdf b/TangDou/LanQiaoBei/2023年12月STEMA C++组中高级真题.pdf
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