HDU_1043Eight

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 1213 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 1213 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

代码：<参考>

#include<iostream>#include<stdio.h>#include<string.h>#include<string>#include<queue>using namespace std;const int MAXN=1000000;//最多是9!/2int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重//         0!1!2!3! 4! 5!  6!  7!   8!    9!bool vis[MAXN];//标记string path[MAXN];//记录路径int cantor(int s[])//康拖展开求该序列的hash值{    int sum=0;    for(int i=0;i<9;i++)    {        int num=0;        for(int j=i+1;j<9;j++)          if(s[j]<s[i])  num++;        sum+=(num*fac[9-i-1]);    }    return sum+1;//0!为1.}struct Node{    int s[9];    int loc;//表示“0”的位置    int status;//表示康拖展开的hash值    string path;//结果路径};int move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//up,down,left,right.char indexs[5]="durl";//和上面的要相反，因为是反向搜索,由目标状态求原始状态int aim=46234;//123456780对应的康拖展开的hash值void bfs(){    memset(vis,false,sizeof(vis));//假设所有的康托展开没有访问过。    Node now,next;    for(int i=0;i<8;i++)//初始化为目标状态now.s[i]=i+1;    now.s[8]=0;    now.loc=8;    now.status=aim;    now.path="";    queue<Node>q;//广度优先搜索用队列    q.push(now);//当前状态入队    path[aim]="";    while(!q.empty())    {        now=q.front();        q.pop();        int x=now.loc/3;//当前数字在方格中的位置。        int y=now.loc%3;        for(int i=0;i<4;i++)//        {            int tx=x+move[i][0];//数字在四个方向上移动后的坐标            int ty=y+move[i][1];            if(tx<0||tx>2||ty<0||ty>2)continue;            next=now;            next.loc=tx*3+ty;//由坐标，求出现在该位置上的数字。            next.s[now.loc]=next.s[next.loc];            next.s[next.loc]=0;            next.status=cantor(next.s);            if(!vis[next.status])//如果该康托展开值没有出现过，即该排序序列没出现过,判断是否重复            {                vis[next.status]=true;//更新访问状态                next.path=indexs[i]+next.path;//一步路径                q.push(next);//现在状态入队                path[next.status]=next.path;//存放每步成功的路径            }        }    }}int main(){    char ch;    Node cur;    bfs();    while(cin>>ch)    {        if(ch=='x') //把“x”看做“0”.{cur.s[0]=0;     cur.loc=0;}        else cur.s[0]=ch-'0';        for(int i=1;i<9;i++)        {            cin>>ch;            if(ch=='x')            {                cur.s[i]=0;                cur.loc=i;            }            else cur.s[i]=ch-'0';        }        cur.status=cantor(cur.s);        if(vis[cur.status])        {            cout<<path[cur.status]<<endl;        }        else cout<<"unsolvable"<<endl;    }return 0;}

思路解析：

经典八码难题，康托展开判重 + BFS反向搜索。

康托展开：实质是计算当前排列在所有由小到大全排列中的顺序，因此是可逆的。

X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[1]*0!

例如，3 5 7 4 1 2 9 6 8 展开为 98884。因为X=2*8!+3*7!+4*6!+2*5!+0*4!+0*3!+2*2!+0*1!+0*0!=98884.

解释：

排列的第一位是3，比3小的数有两个，以这样的数开始的排列有8!个，因此第一项为2*8!

排列的第二位是5，比5小的数有1、2、3、4，由于3已经出现，因此共有3个比5小的数，这样的排列有7!个，因此第二项为3*7!

以此类推，直至0*.

当使用BFS算法时，首先要先检验上、下、左、右的邻居节点,

由于初始状态很多，目标状态只有一种，所以有目标状态反向求初始状态。

网上关于八数码的解题有很多，有人总结了八境界，膜拜了。。。

我个人只是认为这个方法，好理解一点，与大家共享。。。

还有代码的样例输出与结果不一样，估计是打印错误，代码可以AC掉的。。。