最大子矩阵

网址：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=67090#problem/B

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

解题思路： 以i，j为循环层，temp[k]表示从i行到j行第k列的和，然后求temp[]数组序列的最大连续子列的最大和。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <stack>
using namespace std;
#define INF 0x3f3f3f3f
int a[110][110];
int temp[110];

int main()
{
    int n;
    int MAX;
    while(scanf("%d",&n)!=EOF)
    {
        MAX=-1*INF;
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        scanf("%d",&a[i][j]);
        for(int i=0;i<n;i++)
        {
            memset(temp,0,sizeof(temp));
            for(int j=i;j<n;j++)
            {
                for(int k=0;k<n;k++)
                    temp[k]+=a[j][k];//从i行到j行的每一列的和；
                int s=0;
                for(int k=0;k<n;k++)//数组temp[]的最大连续子列的最大和；
                 {
                     if(s<0) s=0;
                     s=s+temp[k];
                     if(MAX<s) MAX=s;
                 }
            }
        }
        printf("%d\n",MAX);
    }
    return 0;
}