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浙大在线评测 1074 To the Max

作者:未知 来源:月光软件站 加入时间:2005-2-28 月光软件站

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].



Input:

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


Output:


15


Solution:

// 声明:本代码仅供学习之用,请不要作为个人的成绩提交。
// http://blog.csdn.net/mskia
// email: bitrain@hotmail.com

#include <iostream.h>

const int N=100;

int array[N][N];

int main( void ) {
    int n;
    cin >> n;
 
    for( int i = 0 ; i < n ; ++i ) {
        for( int j = 0 ; j < n ; ++j ) {
            cin >> array[ i ][ j ] ;
        }
    }

    int sum = maxSum2 ( n );
    cout << sum << endl;
 
    return 0;
}

int maxSum( int n , int *p ) {
    int sum = 0 , b = 0;
    for( int i = 0 ; i < n  ; ++i ) {
        if ( b > 0 ) {
            b += * ( p + i ) ;
        } else {
            b = * ( p + i ) ;
        }

        if ( b > sum ) {
            sum = b;
        }
    }
    
    return sum;
}

int maxSum2( int n ) {
    int sum = 0 ;
    int *b = new int [ n ];
    for( int i = 0 ; i < n ; ++i ) {
        for( int k = 0 ; k < n ; ++k ) {
            *( b + k ) = array [ i ][ k ] ;
        }
  
        for( int j = i + 1 ; j < n ; ++j ) {
            for( int k = 0 ; k < n ; ++k ) {
                *( b + k ) += array [ j ][ k ] ;
            }
        }
        
        int max = maxSum( n , b );
        if( max > sum ) {
           sum = max ;
        }
    }
    
    return sum;
}




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