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1005 B

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#include <bits/stdc++.h>
using namespace std;
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const int N = 150;
int a[N][N];
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int main() {
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int n;
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cin >> n;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++) {
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cin >> a[i][j];
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a[i][j] += a[i - 1][j]; // 按列计算前缀和
}
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int ans = INT_MIN; // 预求最大,先设最小
for (int i = 1; i <= n; i++) // 最底下一行
for (int k = 1; k <= i; k++) { // 往上k行
int dp[N] = {0}; // dp[j]表示压缩的矩形前j列的最大累加值
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for (int j = 1; j <= n; j++) { // 第j列
int s = a[i][j] - a[i - k][j]; // 求压缩的矩形第j列的值
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dp[j] = max(dp[j - 1] + s, s); // 动态规划,到j列为止最大的连续累加和
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ans = max(ans, dp[j]); // 更新答案
}
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}
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cout << ans << endl; // 愉快AC
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return 0;
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}