Problem Description


Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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


本题的题意是给一个数字矩阵，求出这个矩阵中和最大的小矩阵（可以包括这个大矩阵）


这题的思路是遍历所有的矩阵求出最大的和的矩阵  


方法如下：


样例中 0 -2 -7  0


            9  2  -6  2


           -4 1 -4 1


           -1 8 0 -2可以每个数字向右相加变成


           0 -2 -9 -9


           9 11 5 7


           -4 -3 -7 -6


            -1 7 7  5  


 在这个矩阵中每个数字表示的是此行的数相加到这个数的和


每个竖排的数相加就是所关联的数的矩阵


所以遍历所有的竖排的数相加便能得出每个矩阵的和 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 100
int a[N][N];

int  find(int n) 
{
    int i,j,MAX,k,s;
    MAX=-1000;
    
    for(i=1;i<=n;i++)
     for(j=i;j<=n;j++)
     {
        s=0;
        for(k=1;k<=n;k++)
         {
          if(s<=0) s=0;
            s+=a[k][j]-a[k][i-1];
            if(s>MAX)  MAX=s;
          }
     }
    
    return MAX;
}
int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
    memset(a,0,sizeof(a));
    for(i=1;i<=n;i++)
     for(j=1;j<=n;j++)
     {
     cin>>a[i][j];
     a[i][j]+=a[i][j-1];
    }
     printf("%d\n",find(n));
     }
     return 0;
}



从而求出最大的和


代码如下：