hdu 1081 最大子矩阵求和问题

http://acm.hdu.edu.cn/showproblem.php?pid=1081

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
   
   
   
   
   
   
   
   


   
   
   
   

#include <string.h>
#include <stdio.h>
#include <iostream>
using namespace std;
int a[1005][1005];
int dp[1005][1005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
      for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
         {
             cin >>a[i][j];
             dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]+a[i][j];
         }
     /* for(int i=1;i<=n;i++)
      {
         for(int j=1;j<=n;j++)
             cout<< dp[i][j]<< endl;
        cout << endl;
      }*/
      int sum=0,sum1;
      int maxx=-99999;
      for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)
        {
            int sum=0;
            for(int k=1;k<=n;k++)
            {
              sum1=dp[j][k]-dp[i-1][k]-dp[j][k-1]+dp[i-1][k-1];
               if(sum>0)
                  sum+=sum1;
                else
                  sum=sum1;
                if(sum>maxx)
                  maxx=sum;
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}

http://blog.csdn.net/yusiguyuan/article/details/12877103