分类： 广搜
2013-08-13 21:20 
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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
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1. #include<stdio.h>  
2. #include<iostream>  
3. #include<queue>  
4. using namespace std;  
5. typedef struct nn  
6. {  
7.     char way;//记录路径  
8.     int fath;//记录父节点  
9. }node1;  
10. typedef struct nod  
11. {  
12.     int aa[10];  
13.     int n,son;//n为9在aa中的位置  
14. }node2;  
15. int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}},fac[10];  
16. node1 Node[370000];//节点  
17. void set_fac()//计算0到8的阶层  
18. {  
19.     fac[0]=1;  
20.     for(int i=1;i<=8;i++)  
21.     fac[i]=fac[i-1]*i;//printf(“%d”,fac[8]);  
22. }  
23. int cantor(int aa[])//康托展开  
24. {  
25.     int i,j,ans=0,k;  
26.     for(i=0;i<9;i++)  
27.     {  
28.         k=0;  
29.         for(j=i+1;j<9;j++)  
30.         if(aa[i]>aa[j])  
31.         k++;  
32.         ans+=k*fac[8-i];  
33.     }  
34.     return ans;  
35. }  
36. void bfs(int a[])  
37. {  
38.     queue<node2>Q;  
39.     node2 q,p;  
40.     int e,tx,ty,tem,t=0;  
41.     for(e=0;e<9;e++) q.aa[e]=a[e];  
42.     q.n=8;q.son=0;  
43.     Node[q.son].fath=0;//把最终父节点记为0，也就是本身  
44.     Q.push(q);  
45.     while(!Q.empty())  
46.     {  
47.         q=Q.front(); Q.pop();  
48.         for(e=0;e<4;e++)  
49.         {  
50.             p=q;  
51.             tx=q.n%3+dir[e][0];ty=q.n/3+dir[e][1];  
52.             if(tx>=0&&ty>=0&&tx<3&&ty<3)  
53.             {  
54.                 p.n=ty*3+tx;  
55.                 tem=p.aa[p.n];p.aa[p.n]=p.aa[q.n];p.aa[q.n]=tem;  
56.                 p.son=cantor(p.aa);  
57.                 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  
58.                 {  
59.                     Node[p.son].fath=q.son;//当前节点的父节点就是上一个节点  
60.                     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  
61.                     if(e==1)Node[p.son].way=‘r’;  
62.                     if(e==2)Node[p.son].way=‘u’;  
63.                     if(e==3)Node[p.son].way=‘d’;  
64.                     Q.push(p);  
65.                 }  
66.             }  
67.         }  
68.     }  
69. }  
70. int main()  
71. {  
72.     int i,j,s,ss[10],a[10];  
73.     char ch[50] ;  
74.     for(i=0;i<9;i++)//目标  
75.         a[i]=i+1;  
76.     for(i=0;i<370000;i++)  
77.     Node[i].fath=-1;  
78.     set_fac();//计算阶层  
79.         bfs(a);//开始从目标建立一树  
80.   
81.     while(gets(ch)>0)  
82.     {  
83.         for(i=0,j=0;ch[i]!=‘\0’;i++)//把字符串变成数子  
84.         {  
85.              if(ch[i]==‘x’)  
86.             ss[j++]=9;  //把x变为数子9  
87.             else if(ch[i]>=‘0’&&ch[i]<=‘8’)  
88.             ss[j++]=ch[i]-‘0’;  
89.         }  
90.         s=cantor(ss);//算出初态康托值  
91.        if(Node[s].fath==-1) {printf(“unsolvable\n”);continue;}//不能变成目标  
92.          
93.         while(s!=0)  
94.         {  
95.             printf(“%c”,Node[s].way);  
96.             s=Node[s].fath;  
97.         }  
98.         printf(“\n”);  
99.     }  
100. }  
101. /* 
102. 1 2 3 4 5 6 7 8 x 
103.  
104. 2 1  4 3 5 x 6 8 7 
105. unsolvable 
106. 2 1  4 3 5 x 6 8 7 
107. drdlurdruldruuldlurrdd 
108. 8 5 6 4 x 3 4 1 2 
109. rulddruulddluurddrulldrurd 
110. 8 5 6 4 x 3 4 1 2 
111. urdluldrurdldruulddluurddr 
112.  
113. */