FLIP GAME（DFS）

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

解题思路

首先4*4的棋盘，若能反转到满足要求，最多只要15步，最少为0步，所以以步数来深搜；

要注意的是，找到方法后回溯，达到步数后回溯，越出棋盘后回溯；

还有棋盘的转化方式

AC代码

#include<stdio.h>int map[4][4],step,flag=0;void turn(int i,int j)                                      //转换，！运算 或者 ^1运算

{<span style="white-space:pre"></span>    //0^1=1;  1^1=0;    map[i][j]=!map[i][j];    if(i>0)        map[i-1][j]=!map[i-1][j];    if(i<3)        map[i+1][j]=!map[i+1][j];    if(j>0)        map[i][j-1]=!map[i][j-1];    if(j<3)        map[i][j+1]=!map[i][j+1];}int range()                                                 //判定棋子是否全部一样{    int i,j;    for(i=0;i<4;i++)        for(j=0;j<4;j++)            if(map[i][j]!=map[0][0])                return 0;    return 1;}int dfs(int i,int j,int dp){    if(dp == step)    {        flag=range();        return 0;    }    if(flag || i==4)        return 1;                                   //如果已经找到答案，就一直返回到最初dfs,返回到主函数    turn(i,j);    if(j<3)        dfs(i,j+1,dp+1);    else        dfs(i+1,0,dp+1);    turn(i,j);                                     //dp == step 时，如果flag == false 且 未到最后一个棋子    if(j<3)                                        //只回溯到上一层，还原翻转前的状态，向下一个棋子出发        dfs(i,j+1,dp);    else        dfs(i+1,0,dp);}int main(){    int i,j;    char a;    for(i=0;i<4;i++)    {        for(j=0;j<4;j++)        {            scanf("%c",&a);            if(a=='b')                map[i][j]=0;<span style="white-space:pre"></span>//棋盘转换成0和1的二维数组，方便处理            else                map[i][j]=1;        }        getchar();    }    for(step=0;step<16;step++)    {        flag=0;        dfs(0,0,0);        if(flag)        {            printf("%d\n",step);            return 0;        }    }    printf("Impossible\n");    return 0;}