|NOIOJ|动态规划|1768:最大子矩阵

http://noi.openjudge.cn/ch0206/1768/

枚举矩阵起始行和结束行O(n^2), 然后再枚举一维O(n)求最大字段和即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ms(i,j) memset(i, j, sizeof(i));
using namespace std;
int a[105][105];
int n;
int main()
{
scanf("%d", &n);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
{
scanf("%d", &a[i][j]);
}
int ans = -100000000;
for (int i=1;i<=n;i++)//枚举矩阵起始行
{
int b[105];
ms(b,0);
for (int j=i;j<=n;j++)//枚举矩阵结束行
{
for (int k=1;k<=n;k++) b[k] += a[j][k];//累加和
int f = -100000000;
int ret = -100000000;
for (int k=1;k<=n;k++) //动态规划求最大字段和
{f = max(f+b[k], b[k]); ret = max(ret, f);}
ans = max(ans, ret);
}
}
printf("%d\n", ans);
return 0;
}