35 lines
825 B
C++
35 lines
825 B
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int n, m;
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int a[N][N], s[N][N];
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int res;
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/*
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4 5
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1 1 0 0 0
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1 0 1 0 0
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0 0 0 1 1
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0 0 0 1 0
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答案：6
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*/
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int main() {
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cin >> n >> m;
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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 >> a[i][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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}
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for (int x1 = 1; x1 <= n; x1++)
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for (int y1 = 1; y1 <= m; y1++)
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for (int x2 = x1 + 1; x2 <= n; x2++)
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for (int y2 = y1 + 1; y2 <= m; y2++) {
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if (s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] == 0)
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res = max(res, (x2 - x1 + 1) * (y2 - y1 + 1));
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}
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printf("%d\n", res);
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return 0;
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} |