HDU 1081

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5027 Accepted Submission(s): 2388

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

#include<stdio.h>
#include<string.h>
int a[101][101],b[101];
int subsequencesum(int a[],int n)
{
    int sum=0,maxsum=-0x7fffff,i;
    for(i=1;i<=n;i++)
        if(maxsum<a[i])
            maxsum=a[i];
    if(maxsum<=0)
        return maxsum;
    for(i=0;i<n;i++)
    {
        sum+=a[i+1];
        if(sum>maxsum)
            maxsum=sum;   
        else
            if(sum<0)
                sum=0;
    }
    return maxsum;
}                     
int main()
{
         int n,max,ans,temp;
        int i,j,k,T,m;
         while(~scanf("%d",&n))//说的是一组，实际却是多组 
         {
            temp=ans=max=-0x7fffff;
            for(i=1;i<=n;i++)
                for(j=1;j<=n;j++)
                    scanf("%d",&a[i][j]);
            for(i=1;i<=n;i++)
            {                         
                    memset(b,0,sizeof(b));
                    for(j=i;j<=n;j++)
                    {
                             for(k=1;k<=n;k++)
                            {
                                b[k]+=a[j][k];
                            }
                            ans=subsequencesum(b,n);//按行枚举 ,猜测按列枚举时间差不多， 
                            if(temp<ans) 
                                temp=ans;
                    }
            }
            printf("%d\n",temp);
        }
        //while(1);
        return 0;
}