POJ 1050 To the Max

枚举行之间的组合，在求最大连续和。
上学期好像就做过的。。。。。和City Game有种一样的感觉。

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35605		Accepted: 18697

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

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

Sample Output

15

Source

Greater New York 2001

#include <iostream>
#include <cstring>

using namespace std;

int a[110][110];
int b[110];

int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
          cin>>a[i][j];

    int maxn=-999999999;

    for(int i=0;i<n;i++)
    {
       for(int j=0;j<=i;j++)///[i,j]之间的和
       {
           memset(b,0,sizeof(b));
           for(int l=0;l<n;l++)
               for(int k=j;k<=i;k++)
               {
                   b[l]+=a[k][l];
               }
           int sum=0;
           for(int m=0;m<n;m++)
           {
               sum+=b[m];
               maxn=max(maxn,sum);
               if(sum<0) sum=0;
           }
       }
    }

    cout<<maxn<<endl;

    return 0;
}