位元運算-BFS-Flip Game-POJ 1753

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

給十六個格子標上序號

                                        0     1     2     3

                                        4     5     6     7

                                        8     9     10    11

                                        12   13   14    15

便可以採用二進位記錄不同的狀態

                      0    1    0    0    0    1    1    0    0    0    1    1    0    1
   1    1

                     15   14  13  12   11  10  9    8    7    6    5    4    3    2   
1    0

                                                  此時便記錄下了

                         14        10         9         5        4          2          1      0

最後採用十進位儲存此時的狀態 1表示‘b’，0表示‘w’，再做相應的翻轉，既0 -> 1,  1 -> 0

#include<iostream>#include<cstring>using namespace std;char ma[5][5];//用於輸入int states[65535]={0};//（核心部分）選項組因為最大為2^16-1=65535所以數組大小為65535int step[65535]={0};//記錄步數int rear=0,front=0;//記錄頭與尾bool vis[65535];//用於記錄是否搜過//輸入，並記錄初始狀態，放入隊列首位置void shuru(){    int i,state=0;    for(i=0;i<4;i++){        cin>>ma[i];        int j;        for(j=0;j<4;j++)            if(ma[i][j]=='b')                state|=1<<(i*4+j);    }    states[rear++]=state;//記錄初始狀態放入隊列首    vis[state]=true;//記錄已走過}//反轉一個，併產生影響int fanzhuan(int stat,int i){    int state=0;    state|=1<<i;    if(i%4!=0)state|=1<<(i-1);    if((i+1)%4!=0)state|=1<<(i+1);    if(i-4>=0)state|=1<<(i-4);    if(i+4<16)state|=1<<(i+4);    return (state^stat);}//BFS寬度遍曆，搜尋每種狀態，不斷放入隊列中，找到後立即返回真（true）bool bfs(){    while(front<rear)    {        int state=states[front++];//取出隊列頭，記錄此時狀態        if(0==state||65535==state)//一旦找到，立即返回        {            cout<<step[state]<<"\n";            return true;        }        int i;        for(i=0;i<16;i++)//遍曆每種狀態        {            int st;            st=fanzhuan(state,i);            if(!vis[st])//防止重複遍曆            {                states[rear++]=st;//不可以的加到隊列尾部                vis[st]=true;                step[st]=step[states[front-1]]+1;//步數加1            }        }    }    return false;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    shuru();    if(!bfs())cout<<"Impossible\n";    return 0;}