hdu1081 To the max(dp 矩阵压缩)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8947    Accepted Submission(s): 4323

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>
#define N 110
int a[N][N];
int dp(int b[],int n)
{
    int sum=0;
    int max=0;
    for(int i=0;i<n;i++)
    {
        sum+=b[i];
        if(sum<0)
        {
            sum=0;
        }
        max=max>sum?max:sum;
    }
    return max;
}
int main()
{
    int n;
    int b[N];
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        int max=0,sum;
        for(int i=0;i<n;i++)
        {
            memset(b,0,sizeof(b));
            for(int j=i;j<n;j++)
            {
                for(int k=0;k<n;k++)
                {
                    b[k]+=a[j][k];
                }
                sum=dp(b,n);
                max=max>sum?max:sum;
            }
        }
        printf("%d
",max);
    }    
    return 0;
}
//矩阵压缩
//将i~j行压缩成一行，存在b[]中，然后求 数组b[]的最大值。