【原】 POJ 1050 To the Max 求二维矩阵的最大子矩阵 解题报告

 

http://poj.org/problem?id=1050

方法：二维转一维
1、使用积累数组cumarr[1...n][1...n]，使得求任意块儿子矩阵和的复杂度为O(1)，cumarr[i][j]为cumarr[1...i][1...j]子矩阵的和，
    通过cumarr[i][j]=cumarr[i-1][j]+cumarr[i][j-1]-cumarr[i-1][j-1]计算积累数组，复杂度为O(N*N),
2、枚举+dp
    对矩阵的行枚举，确定矩阵区域的上下界imin和imax，此时就可以进行类似一维数组最大和的方法，每个枚举的矩阵区域负责度O(N)。
总复杂度为O(N*N)+(N*N)*O(N)=O(N^3)

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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 <iostream>

using namespace std ;

//只用1...n，以便和积累数组对应
int a[101][101];
int cumarr[101][101];

void run1050()
{
    int n,val ;
    int i,j ;
    int imin,imax ;
    int maxendinghere,maxsofar ;
    int tmp ;

    scanf( "%d", &n );

    //积累数组边界值
    for( i=0 ; i<=n ; ++i )
    {
        cumarr[0][i] = 0 ;
        cumarr[i][0] = 0 ;
    }

    //输入矩阵并求得积累数组
    for( i=1 ; i<=n ; ++i )
    {
        for( j=1 ; j<=n ; ++j )
        {
            scanf( "%d", &(a[i][j]) ) ;
            cumarr[i][j] = cumarr[i-1][j] + cumarr[i][j-1] - cumarr[i-1][j-1] + a[i][j] ;
        }
    }

    //对矩阵的行枚举，确定矩阵区域的上下界imin和imax，
    maxsofar = a[1][1] ;
    for( imin=1 ; imin<=n ; ++imin )
    {
        for( imax=imin ; imax<= n ; ++imax )
        {
            //进行类似一维数组最大和的方法
            maxendinghere = cumarr[imax][1] - cumarr[imin-1][1] ;
            for( j = 2 ; j<=n ; ++j )
            {
                tmp = cumarr[imax][j] - cumarr[imin-1][j] - cumarr[imax][j-1] + cumarr[imin-1][j-1] ;
                maxendinghere = std::max( maxendinghere+tmp , tmp ) ;
                maxsofar = std::max( maxsofar , maxendinghere ) ;
            }
        }
    }

    printf( "%d\n", maxsofar ) ;
}