Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 20993    Accepted Submission(s): 5634

Special Judge

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

 

 

//就是类似九宫格那个游戏，不过这里9个格子大小都相等

//学了比较多的东西，才懂怎么做

这里我用的的是 A*+逆序数剪枝+hash判重 做的

A*其实也好理解，就是 bfs 升级版

这个博客写的很详细  

因为，无论怎么移动，逆序数的奇偶性是不变的，所以用这个能剪枝

最大的问题就是怎么判重了，最多不过 9！种情况么，362880 ，每种情况对应一个数字，用一个 vis[ ]就能判重了,然后hash 判重就好理解了

还有许多方法，推荐一篇博客  有兴趣可以看看

 

A* + hash 判重 + 曼哈顿距离

1 #include 
      
         2 #include 
       
          3 #include 
        
           4 #include 
         
            5 #include 
          
             6 #include 
           
             7 #include 
            
              8 #include 
             
            
           
          
         
        
       
      

 

IDA* + 曼哈顿距离 这个空间就省下来了，改一下，十五数码也能做的，上一个就不行了，内存爆炸

1 #include 
      
         2 #include 
       
          3 #include 
        
           4 #include 
         
            5   6 #define size 3          //这里规定几数码  7   8 int move[4][2]={
    {-1,0},{
    0,-1},{
    0,1},{
    1,0}};//上 左 右 下  置换顺序  9 char op[4]={
    'u','l','r','d'}; 10  11 int map[size][size],map2[size*size],limit,path[100]; 12 int flag,length; 13  14 // 十五数码 的表 15 //int goal[16][2]= {
     {size-1,size-1},{0,0},{0,1},{0,2}, 16 //                  {0,3},{1,0},{1,1},{1,2}, 17 //                  {1,3},{2,0},{2,1},{2,2}, 18 //                  {2,3},{3,0},{3,1},{3,2}};//各个数字应在位置对照表 19  20 // 八数码 的表 21 int goal[9][2]= {
    {size-1,size-1},{
    0,0},{
    0,1},{
    0,2}, 22                                  {
    1,0},{
    1,1},{
    1,2}, 23                                  {
    2,0},{
    2,1}};      //各个数字应在位置对照表 24  25 int nixu(int a[size*size]) 26 { 27     int i,j,ni,w,x,y;  //w代表0的位置 下标，x y 代表0的数组坐标 28     ni=0; 29     for(i=0;i
          
           a[j]) 38                 ni++; 39         } 40     } 41     //x=w/size; 42     //y=w%size; 43     //ni+=abs(x-(size-1))+abs(y-(size-1));  //最后加上0的偏移量 44     if(ni%2==0) 45         return 1; 46     else 47         return 0; 48 } 49  50 int hv(int a[][size])//估价函数，曼哈顿距离，小等于实际总步数 51 { 52     int i,j,cost=0; 53     for(i=0;i
           
            0)//不和上一次移动方向相反，对第二步以后而言105 continue;106 nx=sx+move[i][0]; //移动的四步 上左右下107 ny=sy+move[i][1];108 109 if( 0<=nx && nx