'commit'

TangDou/AcWing/BeiBao/1020_3.cpp

@@ -1,24 +0,0 @@
-#include <bits/stdc++.h>
-using namespace std;
-const int N = 1010;
-const int M = 110;
-int n, m, m1, m2;
-int f[N][M][M];
-
-int main() {
-    cin >> m1 >> m2 >> n;
-    memset(f, 0x3f, sizeof f);
-    f[0][0][0] = 0;
-    
-    for (int i = 1; i <= n; i++) {
-        int u, v, w;
-        cin >> u >> v >> w;
-        for (int j = 0; j <= m1; j++)
-            for (int k = 0; k <= m2; k++) {
-                f[i][j][k] = f[i - 1][j][k];
-                f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - u)][max(0, k - v)] + w);
-            }
-    }
-    cout << f[n][m1][m2] << endl;
-    return 0;
-}

TangDou/AcWing/BeiBao/背包问题专题.md

@@ -110,6 +110,72 @@ int main() {
 - $01$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。
 
+
+**[$AcWing$ $1020$. 潜水员](https://www.acwing.com/problem/content/description/1022/)**
+二维费用背包问题，但是是一道 **反题**，与正常题目相左的一道题。
+人家是装不下就不再装了，它的含义是相反的：我还缺少多少，如果你给的多，那么直接把我装满~
+
+老套路，二维好想，一维好记：
+
+**二维写法**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1010;
+const int M = 110;
+int n, m, m1, m2;
+int f[N][M][M];
+
+int main() {
+    cin >> m1 >> m2 >> n;
+    memset(f, 0x3f, sizeof f);
+    f[0][0][0] = 0;
+
+    for (int i = 1; i <= n; i++) {
+        int v1, v2, w;
+        cin >> v1 >> v2 >> w;
+        for (int j = 0; j <= m1; j++)
+            for (int k = 0; k <= m2; k++) {
+                f[i][j][k] = f[i - 1][j][k];
+                f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - v1)][max(0, k - v2)] + w);
+            }
+    }
+    cout << f[n][m1][m2] << endl;
+    return 0;
+}
+```
+
+**一维写法**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 22, M = 80;
+
+int n, m, K;
+int f[N][M];
+
+int main() {
+    cin >> n >> m >> K;
+
+    memset(f, 0x3f, sizeof f);
+    f[0][0] = 0;
+
+    while (K--) {
+        int v1, v2, w;
+        cin >> v1 >> v2 >> w;
+        for (int i = n; i >= 0; i--)
+            for (int j = m; j >= 0; j--)
+                f[i][j] = min(f[i][j], f[max(0, i - v1)][max(0, j - v2)] + w);
+    }
+
+    cout << f[n][m] << endl;
+
+    return 0;
+}
+```
+
+
 ### 三、$01$背包之恰好装满
 **[$AcWing$ $278$. 数字组合](https://www.acwing.com/problem/content/280/)**