hdu——1081 To The Max

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

 

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求最大子矩阵的和。思路是构造一个数组sum[i][j]，表示第j列从第一行开始到第i行的和。有了这个数组，我们可以枚举子矩阵的上下界，然后构造数组array存储上下界确定时的前i列和。然后找到两列，他们的差最大就行了。

#include<stdio.h>
#include<string.h>

int sum[104][104];
int array[104];
int mat[105][105];

int min(int a,int b)
{
	return a<b?a:b;
}

int max(int a,int b)
{
	return a>b?a:b;
}

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		memset(sum,0,sizeof(sum));
		memset(array,0,sizeof(array));
		int ans=-0x3f3f3f;
		for(int i=1;i<=n;i++)
		  for(int j=1;j<=n;j++)
		  {
  			scanf("%d",&mat[i][j]);
                        sum[i][j]=mat[i][j]+sum[i-1][j];
  		  }
        for(int up=1;up<=n;up++)
          for(int down=up;down<=n;down++)
          {
          	for(int i=1;i<=n;i++)
          	  array[i]=array[i-1]+sum[down][i]-sum[up-1][i];
       	    int mins=0;
       	    for(int i=1;i<=n;i++)
       	    {
    	       	ans=max(ans,array[i]-mins);
    	       	mins=min(mins,array[i]);
   	        }
          }
        printf("%d\n",ans);
	}
	return 0;
}