Eight
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
#include
#include
#include
#include
#include
#include
using namespace std;
int fac[]= {1,1,2,6,24,120,720,5040,40320,362880};
int dir[4][2]= {{-1,0},{1,0},{0,-1},{0,1}};
struct node
{
int s[10];
int x;
int vis;
string sr;
};
int vis[500010],ss[9];
string str[500010];
char index[5]="durl";
int Hash(int a[])
{
int sum=0;
for(int i=1; i<=8; i++)
{
int num=0;
for(int j=0; ja[i])
num++;
sum+=num*fac[i];
}
return sum;
}
void bfs()
{
memset(vis,0,sizeof(vis));
queueQ;
node now,next;
for(int i=0; i<=8; i++)
now.s[i]=i+1;
now.x=8;
now.vis=Hash(now.s);
now.sr="";
str[now.vis]="";
Q.push(now);
while(!Q.empty())
{
now=Q.front();
Q.pop();
int x=now.x/3;
int y=now.x%3;
for(int i=0; i<4; i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(xx>=0&&xx<=2&yy>=0&&yy<=2)
{
next=now;
next.x=xx*3+yy;
next.s[now.x]=next.s[next.x];
next.s[next.x]=9;
next.vis=Hash(next.s);
if(!vis[next.vis])
{
vis[next.vis]=1;
next.sr=index[i]+next.sr;
str[next.vis]=next.sr;
Q.push(next);
}
}
}
}
}
int main()
{
char s[1100];
bfs();//首先从最终状态向各个方向搜索
while(gets(s))
{
int k=0;
for(int i=0; i='0'&&s[i]<='9')
ss[k++]=s[i]-'0';
else if(s[i]=='x')
ss[k++]=9;
}
if(vis[Hash(ss)])//判断该状态是否出现
cout<