POJ 1050-To the Max（最大子矩陣和）

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. 

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 the sum of the maximal sub-rectangle.

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

8  0 -2

15

Greater New York 2001

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
#define MAXN 101
int a[MAXN][MAXN];
int b[MAXN];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    memset(a,0,sizeof(a));
    for(int i=1; i<=n; ++i)
        for(int j=1; j<=n; ++j)
        {
            cin>>a[i][j];
            a[i][j]+=a[i-1][j];//列上的元素依次求和
        }
    /*for(int i=1; i<=n; ++i)
    {
        for(int j=1; j<=n; ++j)
            cout<<a[i][j]<<" ";
        cout<<endl;
    }*/
    int ans=a[1][1];
    for(int i=0; i<=n-1; ++i)//起始行
        for(int j=i+1; j<=n; ++j)//終止行
        {
            memset(b,0,sizeof(b));//第i~j行k列中能取得的最大值
            for(int k=1; k<=n; ++k)//當前列舉的列
            {
                if(b[k-1]>=0) b[k]=b[k-1]+a[j][k]-a[i][k];
                else b[k]=a[j][k]-a[i][k];
                ans=max(ans,b[k]);
            }
        }
    cout<<ans<<endl;
}