Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. 
As an example, the maximal sub-rectangle of the array: 

0 -2 -7 0 
9 2 -6 2 
-4 1 -4 1 
-1 8 0 -2 
is in the lower left corner: 

9 2 
-4 1 
-1 8 
and has a sum of 15. 

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 
9 2 -6 2 
-4 1 -4 1 
-1 8 0 -2 

Sample Output

15

刚开始 在想RMQ。。。死活没想到。。看到题解后竟然是DP。。。

最大连续子序列和的题。。。。

先求每一行的前缀和，然后i，j的横向区间和为dp【k】【i】-dp【k】【j-1】；

把竖列分成n*（n-1）/2个，然后竖向进行最大连续子序列求和。。（分成这么多个子问题，不多不少全包含。）

最大值用ma存，最后输出就行，复杂度O（n^3）;

就当复习DP了。。。。行吧，还是太菜不够灵活。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define inf 0x3f3f3f3f
int dp[101][101];
int main()
{
    int n,m;
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&m);
            dp[i][j]=dp[i][j-1]+m;
        }
   /* for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
            printf(" %d ",dp[i][j]);
        puts("");
    }*/
    int f=0,sum=0,ma=dp[1][1];
    for(int i=0;i<n;i++)
        for(int j=i+1;j<=n;j++)
        {
            sum=0;
            for(int k=1;k<=n;k++)
            {
                f=dp[k][j]-dp[k][i];
              //  printf(" %d +++\n",f);
                sum+=f;
                ma=max(ma,sum);
                if(sum<0)
                    sum=0;
            }
        }
    cout<<ma<<endl;
    return 0;
}

人一我百，人百我万.夕林山寸，寻梦指尖！