POJ 1050 To the Max （最大连续区间和+暴力枚举，水题）

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 99999999
using namespace std;
int a[111][111],b[111];
int fun(int n)           //求连续子序列和的最大值 
{
	int i,sum,ans;
	sum=0;ans=-inf;
	for(i=1;i<=n;i++) {
		sum+=b[i];
		ans=max(ans,sum);
		if(sum<0) sum=0;
	}
	return ans;
}


int main()
{
	int n,i,j,k,ans;
	cin>>n;
	for(i=1;i<=n;i++) {
		for(j=1;j<=n;j++)
		cin>>a[i][j];
	}
	ans=-inf;
	for(i=1;i<=n;i++) {     //暴力每一种矩阵的情况，选择最大值 
		for(j=1;j<=n;j++) b[j]=0;
		for(j=i;j<=n;j++) {
			for(k=1;k<=n;k++) {
				b[k]+=a[j][k];
			}
			ans=max(fun(k),ans);
		}
	}
	cout<<ans<<endl;
	return 0;
}