八数码（康托展开）

#include <iostream>
#include <cstring>
#define LEN 362888              //状态共9!=362880种
using namespace std;

int dir[4][2] = {{1,0}, {-1,0}, {0,1}, {0,-1}};
int visited[LEN];               //状态标记，步骤记录
int dis[LEN];
int state[LEN][9];              //状态池
int goal[9];                    //目标状态
long int factory[]={1,1,2,6,24,120,720,5040,40320,362880};
bool Contor(int str[], int n)
{
long result = 0;
for(int i = 0; i < n; i++)
{
int counted = 0;
for(int j = i+1; j < n; j++)
{
if(str[i] > str[j])             //当前未出现的元素中是排在第几个
++counted;
}
result += counted*factory[n-i-1];
}
if(!visited[result])                    //没有被访问过
{
visited[result] = 1;
return 1;
}
else
return 0;
}
int bfs()
{
int head = 1, tail = 2;     //为了节省空间，不用队列存储各个状态
int i;
while(head < tail)          //枚举完全部状态退出循环
{
//到达目标状态
for(i = 0; i < 9; i++)
{
if(state[head][i] != goal[i])
break;
}
if(i == 9)
return head;

//找到元素0
for(i = 0; i < 9; i++)
{
if(state[head][i] == 0)
break;
}
//转化为二维
int x = i/3;
int y = i%3;
int z = i;

for(i = 0; i < 4; i++)
{
//新的坐标
int nx = x+dir[i][0];
int ny = y+dir[i][1];
//转化为一维
int nz = 3*nx+ny;
if(nx >= 0 && nx < 3 && ny >= 0 && ny < 3)   //未越界
{
for(int j = 0; j < 9; j++)
state[tail][j] = state[head][j];
swap(state[tail][z], state[tail][nz]);
//                for(int i = 0; i < 9; i++)
//                    cout << state[tail][i] ;
//                cout << endl;
dis[tail] = dis[head]+1;
if(Contor(state[tail], 9))               //新状态未出现过
//cout << 1 << endl,
tail++;                              //队列长度+1,相当于入队
}
}
head++;                                          //队列长度-1，相当于出队
}
return 0;

}
int main()
{
for(int i = 0; i < 9; i++)
cin >> state[1][i];             //初始状态
for(int i = 0; i < 9; i++)
cin >> goal[i];                 //目标状态

int num = bfs();
if(num)
cout << dis[num] << endl;
else
cout << "Impossible" << endl;
return 0;
}