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

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 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