Flip Game POJ

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

        这道题的意思是给定一个棋盘上面有黑白棋，翻动一个棋子就会引起自身及其周围四个方向的棋子变成相反的颜色，求最少翻几个棋子才能让棋盘上的棋子颜色相同。

  这道题用深搜，先写两个函数，一个函数用来判断棋盘是否全为一种颜色，另一用来改变颜色。深搜有三个变量 x，y, step,每走到一个棋子 有两个选择 1，翻，2，不翻，如果翻颜色变反，不翻则继续搜下一个，每翻一次step+1，当所有的棋子颜色相同时输出step。

#include<stdio.h>char a[15][15];int b[5][2]= {0,0,1,0,0,1,-1,0,0,-1};int sum,flag,min;int f(){    for(int i=0; i<4; i++)        for(int j=0; j<4; j++)            if(a[i][j]!=a[0][0])                return 0;    return 1;}void turn(int x,int y){    for(int i=0; i<5; i++)    {        int tx=x+b[i][0];        int ty=y+b[i][1];        if(tx<0||ty<0||tx>=4||ty>=4)            continue;        if(a[tx][ty]=='b')            a[tx][ty]='w';        else if(a[tx][ty]=='w')            a[tx][ty]='b';    }    return;}void dfs(int x,int y,int step){    if(f())    {        if(step<min)            min=step;        return;    }    if(x==4)        return;    turn(x,y);    if(y>=3)        dfs(x+1,0,step+1);    else        dfs(x,y+1,step+1);    turn(x,y);    if(y>=3)        dfs(x+1,0,step);    else        dfs(x,y+1,step);}int main(){    flag=0,min=99999999;    for(int i=0; i<4; i++)        scanf("%s",a[i]);    dfs(0,0,0);    if(min>=0&&min<=16)        printf("%d\n",min);    else        printf("Impossible\n");    return 0;}