问题：
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
思路：
这个题也是参考的大佬的，就是从目标状态，去搜索每一种状态，然后把每一种状态和如何走到的路径留下来，
用到了map映照容器，用于储存每种状态和走到的路径。
代码：
#include<map>
#include<queue>
#include<string>
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int sxy,net[4][2]= {0,1,0,-1,1,0,-1,0};
char dis[5]="lrud";
string db[200000];
int step=0;
map<int,int>se;
struct node
{
int xy;
int b;
string ds;
};
void bfs()
{
queue<node>q;
node h,t;
h.xy=8;
h.b=123456789;
h.ds="";
se[h.b]=step++;
db[se[h.b]]="";
q.push(h);
while(!q.empty())
{
h=q.front();
q.pop();
for(int i=0; i<4; i++)
{
t=h;
int tx=h.xy/3+net[i][0];
int ty=h.xy%3+net[i][1];
if(tx<0||ty<0||tx>=3||ty>=3)
continue;
int d=tx*3+ty,k[10];
for(int j=8; j>=0; j--)
{
k[j]=t.b%10;
t.b=t.b/10;
}
int dk=k[d];
k[d]=k[h.xy];
k[h.xy]=dk;
for(int j=0; j<9; j++)
t.b=t.b*10+k[j];
if(se.find(t.b)!=se.end())
continue;
t.ds=dis[i]+t.ds;
se[t.b]=step++;
db[se[t.b]]=t.ds;
t.xy=d;
q.push(t);
}
}
}
int main()
{
char c;
step=0;
bfs();
while(~scanf("%c ",&c))
{
int s=0;
if(c=='x')
s=9;
else
s=c-'0';
for(int i=1; i<8; i++)
{
scanf("%c ",&c);
if(c=='x')
s=s*10+9;
else
s=s*10+(c-'0');
}
scanf("%c",&c);
if(c=='x')
s=s*10+9;
else
s=s*10+(c-'0');
if(se.find(s)!=se.end())
cout<<db[se[s]]<<endl;
else
cout<<"unsolvable"<<endl;
getchar();
}
return 0;
}