Eight（bfs+全排列的哈希函数）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22207		Accepted: 9846		Special Judge

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr

题意：这是一个8数码问题，听着好高端的样子；就是给你一个3*3的矩阵，包括1~8和x;例

1  2  3

x  4  6

7  5  8 问最少需要变换x几步成为

1  2  3
4  5  6
7  8  x 的形式；

这题的关键是找到一个哈希函数，使得矩阵形成的排列与一个自然数一一对应，这里采用的是全排列的哈希函数，另一个就是BFS加打印路径了；

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 #include<iostream>
 using namespace std;

 const int maxn = ;//根据全排列的哈希函数，n+1个数的排列可以对应n个数的多进制形式，这里九个数对应多进制的最大值为9！-1；
 int factorial[] = {,,,,,,,,};
 int pow[] = {,,,,,,,,};
 int head,tail;
 bool vis[maxn];

 struct node
 {
     char status;
     int id,num,pre;
 }que[maxn];

 int hash(int num)//全排列的哈希函数
 {
     int a[],key,i,j,c;
     for(i = ; i < ; i++)
     {
         a[i] = num%;//a数组倒着存的num，所以求逆序数的时候条件是a[j]<a[i];
         num = num/;
     }
     key = ;
     for(i = ; i < ; i++)
     {
         for(j = ,c = ; j < i; j++)
         {
             if(a[j] < a[i])
                 c++;
         }
         key += c*factorial[i];
     }
     return key;
 }

 void change(int num,int a,int b,char status)//a位置上的数和b位置上的数互换；
 {
     int n1,n2;
     n1 = num/pow[a]%;
     n2 = num/pow[b]%;
     num = num - (n1-n2)*pow[a] + (n1-n2)*pow[b];
     int key = hash(num);
     if(!vis[key])
     {
         vis[key] = true;
         que[tail].num = num;
         que[tail].id = b;
         que[tail].status = status;
         que[tail++].pre = head;
     }
 }

 //打印路径
 void print(int head)
 {
     char s[];
     int c = ;
     while(que[head].status != 'k')
     {
         s[c++] = que[head].status;
         head = que[head].pre;
     }
     s[c] = '\0';
     for(int i = c-; i >= ; i--)
     {
         printf("%c",s[i]);
     }
     printf("\n");
 }

 int main()
 {
     char c;
     int num,id,t;

     num = ;
     for(int i = ; i < ; i++)
     {
         cin>>c;
         if(c == 'x')
         {
             t = ;
             id = i;
         }
         else t = c-'';
         num = *num+t;
     }
     bool flag = false;
     memset(vis,false,sizeof(vis));

     head = ;
     tail = ;
     que[].id = id;
     que[].num = num;
     que[].status = 'k';

     while(head < tail)
     {
         num = que[head].num;
         id = que[head].id;
         if(num == )
         {
             flag = true;
             break;
         }
         if(id > )
             change(num,id,id-,'u');

         if(id < )
             change(num,id,id+,'d');

         if(id% != )
             change(num,id,id-,'l');
         if(id% != )
             change(num,id,id+,'r');
         head++;

     }
     if(flag) print(head);
     else printf("unsolvable\n");
     return ;
 }