题意：

就是还原八数码。输出操作。

题目：

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  45  6  7  89 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  45  6  7  8     5  6  7  8     5  6  7  8     5  6  7  89  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  xr->            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

分析：

用广搜打表，根据广搜的性质，第一次出现某种状态，即是最短路

注意，方向问题。最终结果倒着输出

（因为此题用搜索即可做出，网上有康拓展开（利用康拓展开判重。通过康拓展开知一共有362879种状态。用逆序数判断可行解见八数码可行解。然后宽搜。）和A*（A*是一种启发式搜索，g为已花代价，h为估计的剩余代价，而A*是根据f=g+h作为估价函数进行排列，也就是优先选择可能最优的节点进行扩展。）做法，自行去看，我不会QAQ）orz

康拓展开：https://www.xuebuyuan.com/3261380.html

A*：https://blog.csdn.net/acm_cxlove/article/details/7745323

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<map>
#include<string>
using namespace std;
map<int,int>book;
map<int,string>ans;
int c[4][2]= {0,1,1,0,0,-1,-1,0};
struct node
{int x,y;string s;int k[3][3];
};
void dfs()
{queue<node>q;node u,v;u.x=2,u.y=2;int xx=1;for(int i=0; i<3; i++)for(int j=0; j<3; j++)u.k[i][j]=xx++;u.k[2][2]=0;book[123456780]=1;q.push(u);while(!q.empty()){v=q.front();q.pop();for(int i=0; i<4; i++){int yy=1;int dx=v.x+c[i][0];int dy=v.y+c[i][1];if(dx<0||dy<0||dx>=3||dy>=3)continue;for(int o=0; o<3; o++)for(int j=0; j<3; j++){u.k[o][j]=v.k[o][j];}swap(u.k[dx][dy],u.k[v.x][v.y]);for(int o=0; o<3; o++)for(int j=0; j<3; j++){yy=yy*10+u.k[o][j];}if(book[yy])continue;book[yy]=1;u.x=dx;u.y=dy;if(i==0)u.s=v.s+'l';else if(i==1)u.s=v.s+'u';else if(i==2)u.s=v.s+'r';elseu.s=v.s+'d';q.push(u);ans[yy]=u.s;}}
}
int main()
{char w[50];dfs();while(gets(w)){int ant=1;int len=strlen(w);for(int i=0; i<len; i++){if(w[i]>='1'&&w[i]<='8')ant=ant*10+w[i]-'0';else if(w[i]=='x')ant*=10;}if(book[ant]){string m=ans[ant];int l=m.size();for(int i=l-1; i>=0; i--)cout<<m[i];cout<<endl;}elsecout<<"unsolvable"<<endl;}return 0;
}