To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3768    Accepted Submission(s): 1798


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
 

 

Source
Greater New York 2001
 
 
 
#include<stdio.h>
#include<string.h>
#define MAXN 110
int map[110][110];
/*//数组a,长度为n,的最大连续子段和*/
int MaxSubArray(int *a,int n)
{
    int Max=-65535;
    int i,tmp=0;
    for(i=0;i<n;i++)
    {
        if(tmp>0)tmp+=a[i];
        else tmp=a[i];
        if(tmp>Max)  Max=tmp;
    }    
    return Max;    
}    
/*//求 n 行，m 列的矩阵的最大子矩阵和 */
int MaxSubMatrix(int n,int m) 
{
    int Max=-65535;
    int i,j,k;
    int sum;
    int b[MAXN];
    for(i=0;i<n;i++)
    {
        memset(b,0,sizeof(b));
        for(j=i;j<n;j++)
        {
            for(k=0;k<m;k++) b[k]+=map[j][k];
            sum=MaxSubArray(b,m);
            if(sum>Max)Max=sum;
        }    
    }    
    return Max;
}    
int main()
{
    int n;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
          for(j=0;j<n;j++)
            scanf("%d",&map[i][j]);
        printf("%d\n",MaxSubMatrix(n,n));
    }    
    return 0;
}

文章来源:http://www.cnblogs.com/kuangbin/archive/2011/11/24/2261601.html