POJ 1753 Flip Game

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

题意：给你一个4*4的矩阵，要你去翻牌，翻到矩阵上的牌全部是一样的为止，输出所需的最小步数！

思路：因为翻牌时是上下左右一起翻的，所以当你在同一个地方翻2次相当于没翻，而翻奇数次也会翻到原来的样子，所以每格只能翻一次。所以最多翻的次数是16次。再一次搜索去翻，直到第一次翻到所需的状态

AC代码：

#include<iostream>using namespace std;int map[4][4],step,flag=0;void turn(int i,int j){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(){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;}void dfs(int x,int y,int cur){if(cur==step){flag=range();return ;}if(flag||x==4)return ;turn(x,y);if(y<3)        dfs(x,y+1,cur+1);else        dfs(x+1,0,cur+1);turn(x,y);if(y<3)        dfs(x,y+1,cur);else        dfs(x+1,0,cur);return ;}int main(){int i,j;char a;for(i=0;i<4;i++){for(j=0;j<4;j++){cin>>a;if(a=='b')map[i][j]=0;else map[i][j]=1;}}for(step=0;step<=16;step++){flag=0;dfs(0,0,0);if(flag)break;}if(flag)cout<<step<<endl;else cout<<"Impossible"<<endl;return 0;}