POJ - 1050 - To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48416		Accepted: 25627

Description

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

题意：给出一个正方形的矩阵，求在这个正方形内最大的某个矩阵之和。

#include<stdio.h>
#include<string.h>
#define MAXN 105
int main()
{
    int i,j,k,n,t,sum,max;
    int a[MAXN][MAXN];
    while (scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for (i=1;i<=n;++i)
        {
            for (j=1;j<=n;++j)
            {
                scanf("%d",&t);
                a[i][j]=a[i-1][j]+t;
            }
        }
        max=0;
        for (i=1;i<=n;++i)
        {
            for (j=i;j<=n;++j)
            {
                sum=0;
                for (k=1;k<=n;++k)
                {
                    t=a[j][k]-a[i-1][k];
                    sum+=t;
                    if (sum<0) sum=0;
                    if (sum>max) max=sum;
                }
            }
        }
        printf("%d\n",max);
    }
    return 0;
}