From 4377432f8362e841d8cdd12a931bc0fd45792f11 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Fri, 5 Jan 2024 10:21:55 +0800
Subject: [PATCH] 'commit'

---
 TangDou/Topic/HDU1704.cpp          | 36 +++++++++++++++++
 TangDou/Topic/HDU3631.cpp          | 50 +++++++++++++++++++++++
 TangDou/Topic/【Floyd专题】.md | 65 +++++++++++++++++++++++++++---
 3 files changed, 145 insertions(+), 6 deletions(-)
 create mode 100644 TangDou/Topic/HDU1704.cpp
 create mode 100644 TangDou/Topic/HDU3631.cpp

diff --git a/TangDou/Topic/HDU1704.cpp b/TangDou/Topic/HDU1704.cpp
new file mode 100644
index 0000000..e8872fb
--- /dev/null
+++ b/TangDou/Topic/HDU1704.cpp
@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+using namespace std;
+#define inf 0x3f3f3f3f
+const int N = 510;
+int n, m, x, y, ans;
+int g[N][N];
+
+void floyd() {
+    for (int k = 1; k <= n; k++)
+        for (int i = 1; i <= n; i++) {
+            if (!g[i][k]) continue; // floyd优化
+            for (int j = 1; j <= n; j++)
+                g[i][j] |= g[i][k] & g[k][j]; // 通过k传递,或运算
+        }
+}
+int main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        cin >> n >> m;
+        memset(g, 0, sizeof g);
+        while (m--) {
+            cin >> x >> y;
+            g[x][y] = 1; // x<y
+        }
+        // 计算传递闭包
+        floyd();
+
+        ans = 0;
+        for (int i = 1; i <= n; i++) // 统计答案
+            for (int j = i + 1; j <= n; j++)
+                if (!g[i][j] && !g[j][i]) ans++; // 无法确定大小关系
+        cout << ans << endl;
+    }
+    return 0;
+}
\ No newline at end of file
diff --git a/TangDou/Topic/HDU3631.cpp b/TangDou/Topic/HDU3631.cpp
new file mode 100644
index 0000000..16ae0d4
--- /dev/null
+++ b/TangDou/Topic/HDU3631.cpp
@@ -0,0 +1,50 @@
+#include <bits/stdc++.h>
+using namespace std;
+#define inf 0x3f3f3f3f
+const int N = 310;
+int t, n, m, q, u, v, w;
+int mp[N][N];
+bool flag[N]; // 记录是否标记
+
+void floyd(int k) {
+    for (int i = 0; i < n; ++i)
+        for (int j = 0; j < n; ++j)
+            if (mp[i][j] > mp[i][k] + mp[k][j])
+                mp[i][j] = mp[i][k] + mp[k][j];
+}
+int main() {
+    while (cin >> n >> m >> q && n + m + q) {
+        if (t != 0) printf("\n"); // 谜之格式
+        printf("Case %d:\n", ++t);
+        memset(mp, inf, sizeof(mp));
+        memset(flag, false, sizeof(flag));
+        for (int i = 0; i <= n; ++i)
+            mp[i][i] = 0;
+        while (m--) {
+            scanf("%d %d %d", &u, &v, &w);
+            if (w < mp[u][v])
+                mp[u][v] = w;
+        }
+        while (q--) {
+            scanf("%d", &w);
+            if (w == 0) {
+                scanf("%d", &u);
+                if (flag[u])
+                    printf("ERROR! At point %d\n", u);
+                else {
+                    flag[u] = true;
+                    floyd(u);
+                }
+            } else {
+                scanf("%d %d", &u, &v);
+                if (!(flag[u] && flag[v]))
+                    printf("ERROR! At path %d to %d\n", u, v);
+                else if (mp[u][v] == inf)
+                    printf("No such path\n");
+                else
+                    printf("%d\n", mp[u][v]);
+            }
+        }
+    }
+    return 0;
+}
\ No newline at end of file
diff --git a/TangDou/Topic/【Floyd专题】.md b/TangDou/Topic/【Floyd专题】.md
index 5ce48c2..e1ea57c 100644
--- a/TangDou/Topic/【Floyd专题】.md
+++ b/TangDou/Topic/【Floyd专题】.md
@@ -95,7 +95,7 @@ int main() {
 }
 ```
 
-### 四、练习题
+### 四、打印路径
 
 #### [$HDU-1385$ $Minimum$ $Transport$ $Cost$](http://acm.hdu.edu.cn/showproblem.php?pid=1385)
 
@@ -186,7 +186,7 @@ int main() {
 }
 ```
 
-
+### 五、最小环
 #### [$HDU$-$1599$ $find$ $the$ $mincost$ $route$](https://acm.hdu.edu.cn/showproblem.php?pid=1599)
 
 **类型: 最小环**
@@ -266,14 +266,67 @@ signed main() {
     }
 }
 ```
-
+### 六、传递闭包
 #### [$HDU$-$1704$ $Rank$](https://acm.hdu.edu.cn/showproblem.php?pid=1704)
-（传递闭包）
 
+
+**题意**
+给出$M$对胜负关系，胜负关系有传递性（若$A$胜$B$，$B$胜$C$则$A$胜$C$）, **求有多少对不能确定的胜负关系**
+ 
+**解法**：思路很简单，$floyd$ 一遍做传递闭包，然后暴力枚举就行辣，但是竟然会$TLE$，然后上网学了一种新的优化姿势（其实这种优化用处不大，但由于本题是非常稀疏的图，所以$O(N^3)$几乎变成了$O(N^2)$）
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+#define inf 0x3f3f3f3f
+const int N = 510;
+int n, m, x, y, ans;
+int g[N][N];
+
+void floyd() {
+    for (int k = 1; k <= n; k++)
+        for (int i = 1; i <= n; i++) {
+            if (!g[i][k]) continue; // floyd优化
+            for (int j = 1; j <= n; j++)
+                g[i][j] |= g[i][k] & g[k][j]; // 通过k传递,或运算
+        }
+}
+int main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        cin >> n >> m;
+        memset(g, 0, sizeof g);
+        while (m--) {
+            cin >> x >> y;
+            g[x][y] = 1; // x<y
+        }
+        // 计算传递闭包
+        floyd();
+
+        ans = 0;
+        for (int i = 1; i <= n; i++) // 统计答案
+            for (int j = i + 1; j <= n; j++)
+                if (!g[i][j] && !g[j][i]) ans++; // 无法确定大小关系
+        cout << ans << endl;
+    }
+    return 0;
+}
+```
+
+
+### 七、变形
 #### [$HDU$-$3631$ $Shortest$ $Path$](https://acm.hdu.edu.cn/showproblem.php?pid=3631)
 （变形）
 
-**$TODO$**
-据说可以使用$Dijkstra$算法解决，有空可以试试: **[链接](https://blog.csdn.net/K_R_forever/article/details/80525757)**
+**题意**
+有向图求$2$点间的最短路径，要求只能经过被标记的点
+
+**思路**
+由于只能用标记的点去更新，并且又要求任意两点之间的最短距离，显然$floyd$是最合适的。
+这道题要用$floyd$过的话关键就看对于$floyd$的理解了，因为只有标记的点可以走，为了节省时间，我们可以在新标记点的时候以那点为中转点进行一次$floyd$，这就避免了$n^3$的复杂度
+
+
+
 
 https://juejin.cn/post/6935691567696969764
\ No newline at end of file