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
题目分析：
一个N*N的矩阵中的，寻找最大的子矩阵。
主要应用到矩阵压缩的思想，如果某个子矩阵的和最大，我们可以把他们的和压缩为一行，则此时的连续子序列和胃最大的。检查所有的压缩组合，从第一行开始，检查所有的包含此行的往下的行的子矩阵，找出包含此行的子矩阵的最大值，接着检查包含下一行的往下的所有的子矩阵。
代码：
#include<iostream >
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int jz[141][141],dp[141],sum[141],Max;
int N,i,j;
while(~scanf("%d",&N))
{
Max=-1333; ///Max的值每次都要刷新
for( i=0; i<N; i++)
for( j=0; j<N; j++)
scanf("%d",&jz[i][j]);///给矩阵赋值
for(int k=0; k<N; k++)///循环每行都要作为一个起始行
{
memset(dp,0,sizeof(dp));///dp数组刷新
memset(sum,0,sizeof(sum));///sum数组刷新
for(int i=k; i<N; i++)///从当前行开始循环
{
for(int j=0; j<N; j++)///每一列都要考虑
sum[j]+=jz[i][j];///以前到该列的子矩阵加上该列的
dp[0]=sum[0];
for(int h=1; h<N; h++)
{
dp[h]=max(dp[h-1]+sum[h],sum[h]);///以前的和现在的取大值
if(dp[h]>Max)
Max=dp[h];
}
}
}
printf("%d\n",Max);
}
}