poj-1050-To the Max【DP】

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. 

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 the sum of the maximal sub-rectangle.

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

8  0 -2

15

Greater New York 2001

#include<stdio.h>
#include<string.h>
const int inf=-1<<20;
int m[101][101];
int sum[101][101];
int max(int a,int b){
	return a>b?a:b;
}
int main(){
	int n;
	while(~scanf("%d",&n)){
		memset(sum,0,sizeof(sum));
		int i,j,k;
		for(i=1;i<=n;++i){
			for(j=1;j<=n;++j){
				scanf("%d",&m[i][j]);
				sum[i][j]=sum[i-1][j]+m[i][j]; 
			}
		}
		int maxn=-1000;
                for(i=1;i<=n;++i){
        	  int d=-1000;
        	  for(j=i;j<=n;++j){
        		int res=0;
        		for(k=1;k<=n;++k){
        			res=max(res+sum[j][k]-sum[i-1][k],sum[j][k]-sum[i-1][k]);
        			if(res>maxn) maxn=res;
        		}
        	  }
               }

		printf("%d\n",maxn);
	}
	return 0;
}