~题目链接~
http://poj.org/problem?id=1753
可参考大神题解（位运算）：http://www.cnblogs.com/tanhehe/archive/2013/06/11/3131615.html
http://www.cnblogs.com/kuangbin/archive/2011/07/30/2121677.html
http://poj.org/showmessage?message_id=340320
输入
bwwb bbwb bwwb bwww
结果
4
位运算+队列
知识点:
^ 异或，1^1=0,1^0=1
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 65535+10
#include<queue>
using namespace std;
int data,vis[maxn];
struct node
{
int rond,step;
} N;
int sreach()
{
queue<node>Q;
N.rond=data;
N.step=0;
Q.push(N);
vis[N.rond]=1;//标记这种形式的已访问
if(data==0 || data==65535)
return N.step;
while(!Q.empty())
{
for(int i=0; i<16; i++)//从最后一位开始变换，并记录
{
N.rond=Q.front().rond^(1<<i);//此位点翻转
if(i<12)
N.rond^=1<<(i+4);//此位点上侧点翻转
if(i>=4)
N.rond^=1<<(i-4);//此位点下侧点翻转
if(i%4!=0)
N.rond^=1<<(i-1);//此位点左侧点翻转
if(i%4!=3)
N.rond^=1<<(i+1);//此位点右侧点翻转
if(!vis[N.rond])//判断是否被访问过
{
N.step=Q.front().step+1;
Q.push(N);
vis[N.rond]=1;
}
if(N.rond==0 || N.rond==65535)
return N.step;
}
Q.pop();
}
return -1;//不能转换
}
int main()
{
char s;
for(int i=1; i<=4; i++)
{
for(int j=1; j<=4; j++)
{
data*=2;
scanf("%c",&s);
if(s=='b')
data+=1;//转换成数据
}
getchar();
}
int x=sreach();
if(x==-1)
printf("Impossible
");
else
printf("%d
",x);
return 0;
}
附带~测试数据（测试数据过，稳过）
bwbw wwww bbwb bwwb Impossible bwwb bbwb bwwb bwww 4 wwww wwww wwww wwww 0 bbbb bbbb bbbb bbbb 0 bbbb bwbb bbbb bbbb Impossible bwbb bwbb bwbb bbbb Impossible bwbb wwwb bwbb bbbb 1 wwww wwwb wwbb wwwb 1 wwww wwww wwwb wwbb 1 wbwb bwbw wbwb bwbw Impossible bbbb bwwb bwwb bbbb 4 bwwb wbbw wbbw bwwb 4 bbww bbww wwbb wwbb Impossible bbwb bbbw wwbb wwwb Impossible wwwb wwbw wbww wwbw Impossible bbbb wwww wwbb wbbb Impossible bwwb wbwb wbbb wbbb 4 bwbb bwbb bwbw bbbw 5 wbwb bbbb bbww wbbb 6 bbwb bbbb wbwb bbbb 5