poj1050-To the Max-最大子矩阵-dp

一篇讲解最大子矩阵的博客：http://blog.csdn.net/beiyeqingteng/article/details/7056687

[align=center]To the Max[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46737		Accepted: 24741

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

题目大意
求一个n*n的矩阵的最大子矩阵和
代码
#include <cstring>
#include <cstdio>
#include <iostream>
const int maxn=0x3f3f3f3f;
using namespace std;
int n;
int a[110][110],b[110];
int ans=-maxn;
int maxrow()
{
int i;
int nowmax=0,summax=-maxn;
for(i=0;i<n;++i)
{
if(nowmax<=0) nowmax=b[i];//a[i]作为新的起点
else nowmax+=b[i];
if(nowmax>summax) summax =nowmax;
}
return summax;
}
int main()
{

int i,j;
scanf("%d",&n);
for(i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
scanf("%d",&a[i][j]);
}
}

for(i=0;i<n;++i)//哪一行是子矩阵的起始行
{
memset(b,0,sizeof(b));//这里要初始化
for(j=i;j<n;++j)//枚举行，前i到j行的最大子矩阵
{
for(int k=0;k<n;++k)//枚举列
b[k]+=a[j][k];//b表示从第i行到第j行按列相加得到的一行数
int mmax = maxrow();//这一行的最大子段

if(ans<mmax) ans = mmax;
}
}
printf("%d\n",ans);
return 0;

}