HDU1081 To The Max 求子矩阵最大和

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5023    Accepted Submission(s): 2384

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],max[101];
int main()
{
    int maxsum,r,c,m,n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;++i)
            for(j=1;j<=n;++j)
            {
                scanf("%d",&a[i][j]);
                a[i][j]+=a[i-1][j];                                    
            }
        maxsum=a[1][1];
        for(i=0;i<n;++i)//注意这里技巧，从零开始；
        {
            for(j=i+1;j<=n;++j)
            {
                memset(max,0,sizeof(max));
                for(m=1;m<=n;++m)
                {
                    if(max[m-1]>=0)
                        max[m]=max[m-1]+a[j][m]-a[i][m];//a[j][m]-a[i][m]即为第m列中第i+1行到第j行之和
                    else
                        max[m]=a[j][m]-a[i][m];
                    if(maxsum<max[m])
                        maxsum=max[m];
                }
            }
        }
        printf("%d\n",maxsum);
    }
    return 0;
}