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；

  1 #include<stdio.h>

  2 #include<string.h>

  3 #include<queue>

  4 #include<iostream>

  5 using namespace std;

  6 

  7 const int maxn = 362881;//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；

  8 int factorial[9] = {1,1,2,6,24,120,720,5040,40320};

  9 int pow[9] = {100000000,10000000,1000000,100000,10000,1000,100,10,1};

 10 int head,tail;

 11 bool vis[maxn];

 12 

 13 struct node

 14 {

 15     char status;

 16     int id,num,pre;

 17 }que[maxn];

 18 

 19 int hash(int num)//全排列的哈希函数

 20 {

 21     int a[10],key,i,j,c;

 22     for(i = 0; i < 9; i++)

 23     {

 24         a[i] = num%10;//a数组倒着存的num，所以求逆序数的时候条件是a[j]<a[i];

 25         num = num/10;

 26     }

 27     key = 0;

 28     for(i = 1; i < 9; i++)

 29     {

 30         for(j = 0,c = 0; j < i; j++)

 31         {

 32             if(a[j] < a[i])

 33                 c++;

 34         }

 35         key += c*factorial[i];

 36     }

 37     return key;

 38 }

 39 

 40 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；

 41 {

 42     int n1,n2;

 43     n1 = num/pow[a]%10;

 44     n2 = num/pow[b]%10;

 45     num = num - (n1-n2)*pow[a] + (n1-n2)*pow[b];

 46     int key = hash(num);

 47     if(!vis[key])

 48     {

 49         vis[key] = true;

 50         que[tail].num = num;

 51         que[tail].id = b;

 52         que[tail].status = status;

 53         que[tail++].pre = head;

 54     }

 55 }

 56 

 57 //打印路径

 58 void print(int head)

 59 {

 60     char s[100];

 61     int c = 0;

 62     while(que[head].status != 'k')

 63     {

 64         s[c++] = que[head].status;

 65         head = que[head].pre;

 66     }

 67     s[c] = '\0';

 68     for(int i = c-1; i >= 0; i--)

 69     {

 70         printf("%c",s[i]);

 71     }

 72     printf("\n");

 73 }

 74 

 75 int main()

 76 {

 77     char c;

 78     int num,id,t;

 79 

 80     num = 0;

 81     for(int i = 0; i < 9; i++)

 82     {

 83         cin>>c;

 84         if(c == 'x')

 85         {

 86             t = 9;

 87             id = i;

 88         }

 89         else t = c-'0';

 90         num = 10*num+t;

 91     }

 92     bool flag = false;

 93     memset(vis,false,sizeof(vis));

 94 

 95     head = 0;

 96     tail = 1;

 97     que[0].id = id;

 98     que[0].num = num;

 99     que[0].status = 'k';

100 

101     while(head < tail)

102     {

103         num = que[head].num;

104         id = que[head].id;

105         if(num == 123456789)

106         {

107             flag = true;

108             break;

109         }

110         if(id > 2)

111             change(num,id,id-3,'u');

112 

113         if(id < 6)

114             change(num,id,id+3,'d');

115 

116         if(id%3 != 0)

117             change(num,id,id-1,'l');

118         if(id%3 != 2)

119             change(num,id,id+1,'r');

120         head++;

121 

122     }

123     if(flag) print(head);

124     else printf("unsolvable\n");

125     return 0;

126 }

View Code