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]!='\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



*/