Flip Game_位运算

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40455		Accepted: 17573

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

bwwb
bbwb
bwwb
bwww

Sample Output

4

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int t[20]={
    51200,58368,29184,12544,
    35968,20032,10016,4880,
    2248,1252,626,305,
    140,78,39,19,
};//每张牌翻后的值
struct node
{
    int k;
    int step;
};
bool vis[1<<16];
int N=1<<16;
void bfs(int ans)
{
    queue<node>qu;
    node now,next;
    now.k=ans;
    now.step=0;
    qu.push(now);
    vis[now.k]=true;
    while(!qu.empty())
    {
        now=qu.front();
        qu.pop();
        if(now.k==0||now.k==N-1)//要么全为b即为0，全为w即为1<<16-1；
       {
           printf("%d
",now.step);
           return ;
       }
       for(int i=0;i<16;i++)
       {
           next.k=now.k^t[i];
           next.step=now.step+1;
           if(!vis[next.k])
           {
               vis[next.k]=true;
               qu.push(next);
           }
       }
    }
    printf("Impossible
");
}
int main()
{
    char ch[10];
    int ans=0,cnt=15;
    for(int i=1;i<=4;i++)
    {
        scanf("%s",ch);
        for(int j=0;j<4;j++)
        {
            if(ch[j]=='w')//w标记为1，b为0；
                ans|=1<<cnt;//左移cnt位，
//（cnt初始化为15，随循环--，确定1的位置）（总共16位，刚开始1在第一位上所以最多左移15位）
            cnt--;
        }
    }
    memset(vis,false,sizeof(vis));//记录翻过以后二进制码代表的值是否出现过
    bfs(ans);
    return 0;
}