poj 1077-Eight(八数码+逆向bfs打表)

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

解析：八数码问题，这题如果每次从起点搜到终点可能会超时，由于终点都是一样的，所以可以逆向搜索打表，每种状态可以根据数的相对位置对应一个编号（康拓展开）。

代码如下：

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iterator>
#include<utility>
#include<sstream>
#include<iostream>
#include<cmath>
#include<stack>
using namespace std;
const int INF=1000000007;
const double eps=0.00000001;
int dx[]={-1,0,1,0},dy[]={0,-1,0,1}; //方向数组
char dir[]={'d','r','u','l'};        //方向数组对应的相反方向的字符，由于是逆向打表
int fa[363000],fact[10];             //fa[]保存父亲状态的编号，fact[i]=i！
string ans[363000];                  /对应的打印路径
int Cul(int n)
{
    int ret=1;
    for(int st=1;st<=n;st++)  ret*=st;
    return ret;
}
int id(int B[])                    //得到编号
{
    int ret=0;
    for(int i=0;i<9;i++)
    {
        int less=0;
        for(int j=i+1;j<9;j++)  if(B[j]<B[i])  less++;
        ret+=fact[9-1-i]*less;
    }
    return ret;
}
bool in(int x,int y){return x>=0&&x<3&&y>=0&&y<3;}   //判断是否在界内
struct node
{
    int x,y,ID;                                      //坐标，id值，此时的状态数组
    int A[9];
};
int main()
{
    for(int i=0;i<10;i++)  fact[i]=Cul(i);
    for(int i=0;i<363000;i++)  ans[i].clear();
    node st;
    st.x=2,st.y=2,st.ID=0;
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)  st.A[i*3+j]=i*3+j+1;   //终末状态
    memset(fa,-1,sizeof(fa));
    fa[0]=0;
    queue<node> que;
    que.push(st);
    while(!que.empty())
    {
        node now=que.front();  que.pop();
        for(int op=0;op<4;op++)
        {
            node t=now;
            t.x+=dx[op],t.y+=dy[op];
            if(in(t.x,t.y))
            {
                swap(t.A[now.x*3+now.y],t.A[t.x*3+t.y]);  //交换位置
                int I=id(t.A);
                if(fa[I]==-1)                             //没有被访问过
                {
                    t.ID=I;
                    que.push(t);
                    fa[I]=now.ID;
                    ans[I]=dir[op]+ans[now.ID];          // 得到路径
                }
            }
        }
    }
    string S;
    while(getline(cin,S))
    {
        int rear[9],cnt=0;
        istringstream out(S);
        string single;
        while(out>>single)
            if(single=="x")  rear[cnt++]=9;
            else  rear[cnt++]=single[0]-'0';
        int I=id(rear);
        if(fa[I]!=-1)  cout<<ans[I]<<endl;
        else  cout<<"unsolvable"<<endl;
    }
    return 0;
}