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
Difficulty：
1.康拓展开：类似于状态压缩将一个数组映射为一个数。
适用条件：当知道这个数组的个数是一定的。
2.一位数组和二维数组的互换。
http://wenku.baidu.com/link?url=YeNqa-MsEwqOZTvOR__alZS12Xog8OZDlnIXVA5WORuM0lCPu9LmkFrSsbYFBuFVj_h5F0hf6I4NVZD076lwG_xEAmWJft2HtRKWMIWAkBa
思路:将二维数组看成一个状态，再进行bfs,同时要将二维数组先变形为一位数组再根据康拓展开变形为一个数。
打表，反向搜索。
注意：这里有多解，不止一个答案，方向顺序不同，答案不同。
#include <cstdio>
#include <iostream>
#include<cstring>
using namespace std;
struct Node
{
int a[9];
}Q[500000];//转态表示
int n,eid,cnt,flag,x0,ya,z0,tmpid,x,y;
int vis[500000];
Node s,t;
char ss[5];
int dx[]={0,0,-1,1};
int dy[]={1,-1,0,0};
char dir[]="lrdu";//反向搜索，所以反着写。
char di[500000];
int A[9];
int net[500000];
int getid(int *a)//康拓展开
{
int v=0;
for(int i=0;i<=8;i++)
{
int cnt=0;
for(int j=i+1;j<=8;j++)
if(a[i]>a[j])
cnt++;
v+=cnt*A[8-i];
}
return v;
}
void bfs()
{
memset(vis,0,sizeof(vis));
int front=0,rear=0;
Q[rear++]=s;
//net[eid]=-1;
vis[eid]=1;
while(front<rear)
{
Node q=Q[front++];
tmpid=getid(q.a);//一维数组映射为一个数
for(int i=0;i<9;i++)
{
if(q.a[i]==0)
{x0=i/3;
ya=i%3;//一维数组转换为二维数组。
z0=i;
break;
}}
Node add=q;;
for(int i=0;i<4;i++)
{
x=x0+dx[i];
y=ya+dy[i];
if(x<0||x>=3||y<0||y>=3)
continue;
int z=x*3+y;//二维数组转换为一维数组
add.a[z0]=q.a[z];
add.a[z]=0;
int qid=getid(add.a);
if(vis[qid]==0)
{
vis[qid]=1;
net[qid]=tmpid;//转态转移
di[qid]=dir[i];//记录操作
Q[rear++]=add;
}
add=q;
}
//front++;
}
}
int main()
{ A[0]=1;
for(int i=1;i<=8;i++)
A[i]=A[i-1]*i;//康拓展模板
for(int i=0;i<=8;i++)
if(i==8)
s.a[i]=0;
else
s.a[i]=i+1;
eid=getid(s.a);
bfs();
while(~scanf("%s",ss))
{
if(ss[0]=='x')
s.a[0]=0;
else
s.a[0]=ss[0]-'0';
for(int i=1;i<=8;i++)
{
scanf("%s",ss);//可过滤空格
if(ss[0]=='x')
s.a[i]=0;
else
s.a[i]=ss[0]-'0';
}
int sid=getid(s.a);
if(vis[sid])
{
while(sid!=eid)//上面打的表中推出一步一步的转态转换。
{
putchar(di[sid]);
sid=net[sid];
}
}
else
printf("unsolvable");
puts("");
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。