You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
using namespace std;
|
|
|
|
|
const int N = 150;
|
|
|
|
|
int a[N][N];
|
|
|
|
|
|
|
|
|
|
int main() {
|
|
|
|
|
int n;
|
|
|
|
|
cin >> n;
|
|
|
|
|
for (int i = 1; i <= n; i++)
|
|
|
|
|
for (int j = 1; j <= n; j++) {
|
|
|
|
|
cin >> a[i][j];
|
|
|
|
|
a[i][j] += a[i - 1][j]; // 按列计算前缀和
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
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
|
|
|
|
|
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
|
|
|
|
|
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
|
|
|
|
|
ans = max(ans, dp[j]); // 更新答案
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
cout << ans << endl; // 愉快AC
|
|
|
|
|
return 0;
|
|
|
|
|
}
|