POJ 1753 Flip Game（bfs+状态压缩+位运算）

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

题解：

每次你可以选择一个棋子，他周围上下左右和自己都会颜色反转，问你最少多少步可以把整个棋盘搞为一种颜色

思路：

因为这个棋盘只是4x4比较小，我们之间用2进制的16位数来表示棋盘的状态进行bfs，如果这种状态搜索过了就不放进队列，对于反转的话就是位运算了，就是异或运算，这个要自己体会比较巧妙，如果当状态为0或者状态为2的16次方也就是65535的时候为全部同种颜色可以跳出搜索输出步数

代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define INF 1008611111#define ll long long#define eps 1e-15struct node{    int z;//当前棋盘状态    int ans;//当前步数};queue<node>q;int dirx[4]={0,0,-1,1};//四个方向int diry[4]={-1,1,0,0};int p[65540*2];//2的16次方，防止炸开2的17次int maxx;void bfs(){    int i,j,x,y,z,k,t,pos;    node now,next;    while(!q.empty())    {        now=q.front();        q.pop();        for(i=0;i<4;i++)        {            for(j=0;j<4;j++)            {                t=now.z;                pos=4*i+j;//棋子编号                t^=1<<(15-pos);//处理当前棋子状态                for(k=0;k<4;k++)                {                    x=i+dirx[k];                    y=j+diry[k];                    if(x>=0&&y>=0&&x<4&&y<4)//如果合法                    {                        pos=4*x+y;                        t^=1<<(15-pos);                    }                }                next.ans=now.ans+1;                next.z=t;                if(!p[t])                {                    p[t]=1;                    q.push(next);                    if(p[0]||p[65535])//如果搜到了                    {                        maxx=next.ans;                        return;                    }                }            }        }    }}int main(){    int i,j,k,pos;    char s[10][10];    node now;    now.ans=0;    now.z=0;    for(i=0;i<4;i++)    {        scanf("%s",s[i]);        for(j=0;j<4;j++)        {            if(s[i][j]=='b')            {                pos=4*i+j;//当前棋子编号                now.z|=(1<<pos);//或运算，相当于在二进制的这个位置放一个1            }        }    }    if(now.z==0||now.z==65535)    {        printf("0\n");        return 0;    }    memset(p,0,sizeof(p));    p[now.z]=1;    q.push(now);    maxx=-1;    bfs();    if(maxx==-1)        printf("Impossible\n");    else        printf("%d\n",maxx);    return 0;}