poj 1050最大子矩阵

题目：

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49846		Accepted: 26420

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

分析：

不需要二维树状数组 因为没有修改操作  只要维护一个前缀和就可以了

具体参：http://blog.pureisle.net/archives/266.html 讲得很好

代码：

#include<iostream>
#include<cstring>
using namespace std;
#define N 110
int a

,f
;
int main(){
int n,r;
cin>>r;
for(int i=1;i<=r;++i){
for(int j=1;j<=r;++j){
cin>>a[i][j];
a[i][j]+=a[i-1][j];///a[i][j]代表前i行 第j列的部分和
}
}
int ans=a[1][1];
for(int i=0;i<=r-1;++i){///起始行
for(int j=i+1;j<=r;++j){///终止行
///相当于每次对[i+1,j]行进行操作，将这几行捆绑在一起进行考虑
///a[j][k]-a[i][k]即为[i+1,j]行第k列的部分和
memset(f,0,sizeof(f));
for(int k=1;k<=r;++k){///转换为一维最大子串和 a[j][k]-a[i][k]相当于一维数组里面每次判断的量
if(f[k-1]>=0) f[k]=f[k-1]+a[j][k]-a[i][k];
else f[k]=a[j][k]-a[i][k];
if(ans<f[k]) ans=f[k];
}
}
}
cout<<ans<<endl;
return 0;
}