hdu 1081 To The Max 【最大子矩阵和】

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8882 Accepted Submission(s): 4288

Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

Source
Greater New York 2001

代码：

#include<stdio.h>  
#include<iostream>  
#include<math.h>  
#include<stdlib.h>  
#include<ctype.h>  
#include<algorithm>  
#include<vector>  
#include<string.h>  
#include<queue>  
#include<stack>  
#include<set>  
#include<map>  
#include<sstream>  
#include<time.h>  
#include<malloc.h>  

using namespace std;  

const int MAXN = 1010;
int n;
int p[MAXN][MAXN];

int longmax(int a[],int n)
{
    int b = 0;
    int ans = -10000000;
    for(int i=0;i<n;i++)
    {
        if (b>0)
            b+= a[i];
        else b= a[i];
        ans = max (ans,b);
    }
    return ans;
}

int work()
{
    int t[1010];
    int ans = -10000000;
    for(int i=0;i<n;i++)
    {
        memset(t,0,sizeof(t));
        for(int j = i;j<n;j++)//枚举第i到第j行的所有可能矩阵的和
        {
            int k;
            for(k=0;k<n;k++)
            {
                t[k] += p[j][k];
            }
            int tt = longmax(t,k);
            ans = max(ans,tt);
        }
    }
    return ans;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                scanf("%d",&p[i][j]);
            }
            int ans = work();
            printf("%d\n",ans);
    }
    return 0;
}