hdu1043Eight (经典的八数码)（康托展开+BFS）

建议先学会用康托展开：http://blog.csdn.net/u010372095/article/details/9904497

Problem Description

The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:

 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.

 

Input

You will receive, several descriptions of configuration of the 8 puzzle. One description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle

1 2 3
x 4 6
7 5 8

is described by this list:

1 2 3 x 4 6 7 5 8

 

Output

You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line. Do not print a blank line between cases.

 

Sample Input

2 3 4 1 5 x 7 6 8

 

Sample Output

ullddrurdllurdruldr

#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]!='';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
");continue;}//不能变成目标
       
        while(s!=0)
        {
            printf("%c",Node[s].way);
            s=Node[s].fath;
        }
        printf("
");
    }
}
/*
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

*/