1074 To the Max

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To the Max

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Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 4133   Accepted Submit: 2006  

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Problem

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.

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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15

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Problem Source: Greater New York 2001

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  http://acm.zju.edu.cn/show_problem.php?pid=1074

 

#include  < iostream >
using   namespace  std;

int  matrix[ 101 ][ 101 ];

int  maxSubSum(  int  n,  int  a[] );
int  maxSum(  int  n );

int  main()
... {
    int n;
    
    while ( cin >> n )
    ...{
        for ( int i = 0; i < n; i++ )
        ...{
            for ( int j = 0; j < n; j++ )
            ...{
                cin >> matrix[i][j];
            }
        }
        
        cout << maxSum( n ) << endl;

    }
    
    return 0;
}

int  maxSubSum(  int  n,  int  a[] )
... {
    int max = -127;
    int thisSum = 0;
    
    for ( int i = 0; i < n; i++ )
    ...{
        if ( thisSum > 0 )
            thisSum += a[i];
        else
            thisSum = a[i];
        
        if ( thisSum > max )
            max = thisSum;
    }
    
    return max;
}

int  maxSum(  int  n )
... {
    int b[101];
    int sum = 0;
    int max = -127;
    
    for ( int i = 0; i < n; i++ )
    ...{
        for ( int k = 0; k < n; k++ ) b[k] = matrix[i][k];
        
        sum = maxSubSum( n, b );
        if ( sum > max ) max = sum;
        
        for ( int j = i+1; j < n; j++ )
        ...{
            for ( int k = 0; k < n; k++ ) b[k] += matrix[j][k];
            sum = maxSubSum( n, b );
            if ( sum > max ) max = sum;
        }
    }
    
    return max;
}