poj1681--Painter's Problem(高斯消元问题4)

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Description

There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob‘s brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow. 

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a ‘w‘ to express a white brick while a ‘y‘ to express a yellow brick.

Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can‘t paint all the bricks yellow, print ‘inf‘.

Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww

Sample Output

0
15

题目和http://blog.csdn.net/winddreams/article/details/43152305黑白棋完全一样，只是加了n的大小和多组

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define INF 0x3f3f3f3f
int Map[400][400] , a[400] , freex[400] , x[400] ;
char str[20][20] ;
void init(int n)
{
    int i , j , k ;
    memset(Map,0,sizeof(Map)) ;
    for(i = 0 ; i < n ; i++)
    {
        for(j = 0 ; j < n ; j++)
        {
            a[i*n+j] = str[i][j] - '0' ;
            Map[i*n+j][i*n+j] = 1 ;
            if( i > 0 ) Map[i*n+j][i*n+j-n] = 1 ;
            if( i < n-1 ) Map[i*n+j][i*n+j+n] = 1 ;
            if( j > 0 ) Map[i*n+j][i*n+j-1] = 1 ;
            if( j < n-1 ) Map[i*n+j][i*n+j+1] = 1 ;
        }
    }
    return ;
}
void swap1(int p,int q,int n)
{
    int i , temp ;
    temp = a[p] ;
    a[p] = a[q] ;
    a[q] = temp ;
    for(i = 0 ; i < n ; i++)
    {
        temp = Map[p][i] ;
        Map[p][i] = Map[q][i] ;
        Map[q][i] = temp ;
    }
    return ;
}
int solve(int n)
{
    int i , j , k , t = 0 , num1 = 0 ;
    for(i = 0 ; i < n && t < n ; i++ , t++)
    {
        for(j = i ; j < n ; j++)
            if( Map[j][t] ) break ;
        if( j == n )
        {
            i-- ;
            freex[num1++] = t ;
            continue ;
        }
        if( i != j )
            swap1(i,j,n) ;
        for(j = i+1 ; j < n ; j++)
        {
            if( Map[j][t] )
            {
                a[j] ^= a[i] ;
                for(k = t ; k < n ; k++)
                    Map[j][k] ^= Map[i][k] ;
            }
        }
    }
    for( ; i < n ; i++)
        if( a[i] ) return -1 ;
    if( num1 > 0 ) return num1 ;
    for(i = n-1 ; i >= 0 ; i--)
    {
        x[i] = a[i] ;
        for(j = i+1 ; j < n ; j++)
            x[i] ^= (Map[i][j]*x[j]) ;
    }
    return num1 ;
}
int main()
{
    int t , n , i , j , k , min1 , ans , key ;
    scanf("%d", &t) ;
    while( t-- )
    {
        scanf("%d", &n) ;
        for(i = 0 ; i < n ; i++)
        {
            scanf("%s", str[i]) ;
            for(j = 0 ; j < n ; j++)
            {
                if( str[i][j] == 'w' )
                    str[i][j] = '1' ;
                else
                    str[i][j] = '0' ;
            }
        }
        init(n) ;
        key = solve(n*n) ;
        min1 = INF ;
        if( key == 0 )
        {
            ans = 0 ;
            for(i = 0 ; i < n*n ; i++)
                ans += x[i] ;
            min1 = min( min1 , ans ) ;
        }
        else if( key > 0 )
        {
            int t , temp = 1<<key ;
            for(t = 0 ; t < temp ; t++)
            {
                ans = 0 ;
                memset(x,0,sizeof(x)) ;
                for(j = 0 ; j < key ; j++)
                    if( t & (1<<j) )
                    {
                        x[ freex[j] ] = 1 ;
                        ans++ ;
                    }
                for(i = n*n-1 ; i >= 0 ; i--)
                {
                    for(k = 0 ; k < n*n ; k++)
                        if( Map[i][k] )
                            break ;
                    x[k] = a[i] ;
                    for(j = k+1 ; j < n*n ; j++)
                        x[k] ^= ( Map[i][j]*x[j] ) ;
                    ans += x[k] ;
                }
                min1 = min( ans , min1 ) ;
            }
        }
        if( min1 == INF )
            printf("inf\n") ;
        else
            printf("%d\n", min1) ;
    }
    return 0;
}

poj1681--Painter's Problem(高斯消元问题4)

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原文地址：http://blog.csdn.net/winddreams/article/details/43153621