提供一个 $O(n^3)$ 做法。

#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int sum[N][N], f[N], a[N][N], n, ans;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++)
			cin >> a[i][j], sum[i][j] = sum[i - 1][j] + a[i][j];
	for (int i = 1; i <= n; i ++)
		for (int j = 0; j < i; j ++) 
			for (int k = 1; k <= n; k ++)
				ans = max(ans, f[k] = max(f[k - 1], 0) + sum[i][k] - sum[j][k]);
	cout << ans << '\n';
	return 0;
}