Description

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it‘s black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules: 

1. Choose any one of the 16 pieces. 
   
2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

Consider the following position as an example: 

bwbw 
wwww 
bbwb 
bwwb 
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become: 

bwbw 
bwww 
wwwb 
wwwb 
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal. 

Input

The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

Output

Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it‘s impossible to achieve the goal, then write the word "Impossible" (without quotes).

Sample Input

bwwbbbwbbwwbbwww

Sample Output

4

黑白棋翻转问题，搜索，枚举都可以做，这次用它来练习高斯消元做法

每个有关系的系数为1，每个棋子是否翻转确定了结果是0还是1，得到了异或方程组，使用高斯消元，找出自由元，然后转化为二进制后，枚举自由元，找出最优解。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define INF 0x3f3f3f3fint Map[20][20] , a[20] , min1 , x[20] , freex[20] ;char str[5][5] ;void getMap(int k){    int i , j ;    memset(Map,0,sizeof(Map)) ;    for(i = 0 ; i < 4 ; i++)    {        for(j = 0 ; j < 4 ; j++)        {            a[i*4+j] = (str[i][j] - '0') ^ k ;            Map[i*4+j][i*4+j] = 1 ;            if( i > 0 )                Map[i*4+j][i*4+j-4] = 1 ;            if( i < 3 )                Map[i*4+j][i*4+j+4] = 1 ;            if( j > 0 )                Map[i*4+j][i*4+j-1] = 1 ;            if( j < 3 )                Map[i*4+j][i*4+j+1] = 1 ;        }    }    return ;}void swap1(int p,int q){    int j , temp ;    temp = a[p] ;    a[p] = a[q] ;    a[q] = temp ;    for(j = 0 ; j < 16 ; j++)    {        temp = Map[p][j] ;        Map[p][j] = Map[q][j] ;        Map[q][j] = temp ;    }    return ;}int solve(){    int i , j , k , t = 0 , num1 = 0 ;    for(i = 0 ; i < 16 && t < 16 ; t++ , i++)    {        for(j = i ; j < 16 ; j++)            if( Map[j][t] )                break ;        if( j == 16 )        {            freex[num1++] = t ;            i-- ;            continue ;        }        if( i != j )            swap1(i,j) ;        for(j = i+1 ; j < 16 ; j++)        {            if( Map[j][t] )            {                a[j] = a[j]^a[i] ;                for(k = t ; k < 16 ; k++)                    Map[j][k] = Map[j][k] ^ Map[i][k] ;            }        }    }    for( ; i < 16 ; i++)        if( a[i] )            return -1 ;    if( num1 ) return num1 ;    for(i = 15 ; i >= 0 ; i--)    {        x[i] = a[i] ;        for(j = i+1 ; j < 16 ; j++)            x[i] = x[i]^(Map[i][j]*x[j]) ;    }    return num1 ;}int f(int s){    int i , j , k , key , ans , min1 = INF ;    getMap(s) ;    key = solve() ;    ans = 0 ;    if( key == 0 )    {        ans = 0 ;        for(i = 0 ; i < 16 ; i++)            ans += x[i] ;        min1 = min(min1,ans) ;    }    else if( key > 0 )    {        int temp = 1<<key ;        for(int t = 0 ; t < temp ; t++)        {            memset(x,0,sizeof(x)) ;            ans = 0 ;            for(j = 0 ; j < key ; j++)                if( t & (1<<j) )                {                    x[ freex[j] ] = 1 ;                    ans++ ;                }            for(i = 15 ; i >= 0 ; i--)            {                for(k = 0 ; k < 16 ; k++)                    if( Map[i][k] )                        break ;                x[k] = a[i] ;                for(j = k+1 ; j < 16 ; j++)                    x[k] ^= (Map[i][j]*x[j]) ;                ans += x[k] ;            }            min1 = min(min1,ans) ;        }    }    return min1 ;}int main(){    int i , j , min1 = INF , k , ans , temp ;    for(i = 0 ; i < 4 ; i++)    {        scanf("%s", str[i]) ;        for(j = 0 ; j < 4 ; j++)        {            if( str[i][j] == 'b' )                str[i][j] = '0' ;            else                str[i][j] = '1' ;        }    }    ans = f(0) ;    min1 = min(min1,ans) ;    ans = f(1) ;    min1 = min(min1,ans) ;    if( min1 == INF )        printf("Impossible\n") ;    else        printf("%d\n", min1) ;    return 0;}

poj1753--Flip Game(高斯消元问题2，枚举自由元的首杀)