44 lines
1.1 KiB
C++
44 lines
1.1 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int a[N][N]; // 原始数组
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int s[N][N]; // 二维前缀和数组
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int get(int x1, int y1, int x2, int y2) {
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return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
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}
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/*
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4 5 5
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00000
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01111
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00011
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00011
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参考答案：6
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*/
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int main() {
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int n, m, k;
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cin >> n >> m >> k;
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char c;
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// 输入原始数组
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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cin >> c, a[i][j] = c - '0';
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// 构建二维前缀和数组
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
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// 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
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int ans = 1e9;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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for (int x = i; x <= n; x++)
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for (int y = j; y <= m; y++)
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if (get(i, j, x, y) >= k)
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ans = min(ans, (x - i + 1) * (y - j + 1));
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if(ans==1e9) ans=0;
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cout << ans << endl;
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return 0;
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}
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