'commit'

GESP/Level5/GESP202406/20240601.cpp

@@ -0,0 +1,44 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 110;
+int a[N][N]; // 原始数组
+int s[N][N]; // 二维前缀和数组
+
+int get(int x1, int y1, int x2, int y2) {
+  return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+}
+/*
+4 5 5
+00000
+01111
+00011
+00011
+
+参考答案：6
+*/
+int main() {
+  int n, m, k;
+  cin >> n >> m >> k;
+  char c;
+  // 输入原始数组
+  for (int i = 1; i <= n; i++)
+    for (int j = 1; j <= m; j++)
+      cin >> c, a[i][j] = c - '0';
+  // 构建二维前缀和数组
+  for (int i = 1; i <= n; i++)
+    for (int j = 1; j <= m; j++)
+      s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+
+  // 遍历每个可能的矩形，计算此矩形的二维前缀和，要求大于等k,求得最小的长乘以宽的值，也就是最小矩形大小
+  int ans = 1e9;
+  for (int i = 1; i <= n; i++)
+    for (int j = 1; j <= m; j++)
+      for (int x = i; x <= n; x++)
+        for (int y = j; y <= m; y++)
+          if (get(i, j, x, y) >= k)
+            ans = min(ans, (x - i + 1) * (y - j + 1));
+
+  if(ans==1e9) ans=0;
+  cout << ans << endl;
+  return 0;
+}