# hdu1043Eight (经典的八数码)（康托展开+BFS）

2013-08-13 21:20
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BFS

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
[cpp]
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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. */
原文作者：BFS
原文地址: https://blog.csdn.net/pi9nc/article/details/27580893
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