hdu 1081To The Max

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

题意：求子矩阵的和的最大值

思路：把多维转化为一维，只要会一维的就简单了。。。

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4842    Accepted Submission(s): 2289

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<stdio.h>
#include<string.h>
#define Max(x,y) x>y?x:y
#define INF -0x7fffffff/2
int max_sub(int x[],int n)
{
    int max=x[0];
    int sum=0;
    int i;
    for(i=0;i<n;i++)
    {
        sum+=x[i];
        if(sum<0)
        {
            sum=0;
        }
        if(max<sum)
        {
            max=sum;
        }
    }
    
    return max;
}
int  main()
{
    int t;
    int i,j,k;
    int aa[105][105];
    int bb[105];
    int maxx;

    while(~scanf("%d",&t))
    {
        maxx=INF;
        for(i=0;i<t;i++)
        {
            for(j=0;j<t;j++)
            {
                scanf("%d",&aa[i][j]);
            }
        }
        
        for(i=0;i<t;i++)
        {
            memset(bb,0,sizeof(bb));
            for(j=i;j<t;j++)
            {
                for(k=0;k<t;k++)
                {
                    bb[k]+=aa[j][k];
                }
                
                maxx=maxx>max_sub(bb,t)?maxx:max_sub(bb,t);
            }
        }
        printf("%d\n",maxx);
    }
}