1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
typedef struct nn
{
char way;//记录路径
int fath;//记录父节点
}node1;
typedef struct nod
{
int aa[10];
int n,son;//n为9在aa中的位置
}node2;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}},fac[10];
node1 Node[370000];//节点
void set_fac()//计算0到8的阶层
{
fac[0]=1;
for(int i=1;i<=8;i++)
fac[i]=fac[i-1]*i;//printf("%d",fac[8]);
}
int cantor(int aa[])//康托展开
{
int i,j,ans=0,k;
for(i=0;i<9;i++)
{
k=0;
for(j=i+1;j<9;j++)
if(aa[i]>aa[j])
k++;
ans+=k*fac[8-i];
}
return ans;
}
void bfs(int a[])
{
queue<node2>Q;
node2 q,p;
int e,tx,ty,tem,t=0;
for(e=0;e<9;e++) q.aa[e]=a[e];
q.n=8;q.son=0;
Node[q.son].fath=0;//把最终父节点记为0，也就是本身
Q.push(q);
while(!Q.empty())
{
q=Q.front(); Q.pop();
for(e=0;e<4;e++)
{
p=q;
tx=q.n%3+dir[e][0];ty=q.n/3+dir[e][1];
if(tx>=0&&ty>=0&&tx<3&&ty<3)
{
p.n=ty*3+tx;
tem=p.aa[p.n];p.aa[p.n]=p.aa[q.n];p.aa[q.n]=tem;
p.son=cantor(p.aa);
if(Node[p.son].fath==-1)//为－1时表示这个点没有访问过，那么放入队列
{
Node[p.son].fath=q.son;//当前节点的父节点就是上一个节点
if(e==0)Node[p.son].way='l';//一定要注意了，e=0是向右走，但我们是要往回搜，所以为了在输出时不用再进行转换，直接记录相反的方向
if(e==1)Node[p.son].way='r';
if(e==2)Node[p.son].way='u';
if(e==3)Node[p.son].way='d';
Q.push(p);
}
}
}
}
}
int main()
{
int i,j,s,ss[10],a[10];
char ch[50] ;
for(i=0;i<9;i++)//目标
a[i]=i+1;
for(i=0;i<370000;i++)
Node[i].fath=-1;
set_fac();//计算阶层
bfs(a);//开始从目标建立一树 while(gets(ch)>0)
{
for(i=0,j=0;ch[i]!='\0';i++)//把字符串变成数子
{
if(ch[i]=='x')
ss[j++]=9; //把x变为数子9
else if(ch[i]>='0'&&ch[i]<='8')
ss[j++]=ch[i]-'0';
}
s=cantor(ss);//算出初态康托值
if(Node[s].fath==-1) {printf("unsolvable\n");continue;}//不能变成目标 while(s!=0)
{
printf("%c",Node[s].way);
s=Node[s].fath;
}
printf("\n");
}
}
/*
1 2 3 4 5 6 7 8 x 2 1 4 3 5 x 6 8 7
unsolvable
2 1 4 3 5 x 6 8 7
drdlurdruldruuldlurrdd
8 5 6 4 x 3 4 1 2
rulddruulddluurddrulldrurd
8 5 6 4 x 3 4 1 2
urdluldrurdldruulddluurddr */