POJ 1077 8数码问题（BFS）

Eight

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) :    Accepted Submission(s) : 

Special Judge

Problem Description

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 a description of a configuration of the 8 puzzle. The 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.

 

Sample Input

2 3 4 1 5 x 7 6 8

 

Sample Output

ullddrurdllurdruldr

 

Source

PKU

 

整个程序由以下部分组成：结点，cantor展开（当HASH函数用的），移动函数，BFS，输出函数

由于有调试的部分在里面而且移动函数也不是特别简洁所以显得特别长，BFS效率低下也只有在弱数据的POJ 1077可以过

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <stdio.h>

#define MAX 500000

char str[400000];
    int tot=0;

using namespace std;

struct Note
{
    char s[9];//数码串
    int num;//第多少号
    int fa;//父结点是多少号
    char dir;//从父结点MOVE方向
}note[MAX];

queue<Note> q;

int vis[MAX];

int cantor(char* s)
{
    int len=9;

    int sum=0;
    for(int i=0;i<len;i++)
    {
        int a=0;
        int b=1;
        for(int j=i+1;j<len;j++)
            if(s[j]<s[i])
            {
                a=a+1;
            }
        //sum+=a*(8-i)!
        for(int k=1;k<len-i;k++)
        {
            b*=k;
        }

        sum+=a*b;
    }

    return sum;

}

int move4(char *s,int n)
{
    //会对S进行改变
    //返回1表示进行了移动，返回0表示没有移动
    int pos;
    int len=9;
    if(n==1)//move up
    {
        for(int i=0;i<len;i++)
        {
            if(s[i]=='9')
            {
                pos=i;
                break;
            }
        }
        if(pos-3>=0)
        {
            s[pos]=s[pos-3]^s[pos];
            s[pos-3]=s[pos]^s[pos-3];
            s[pos]=s[pos]^s[pos-3];

            return 1;
        }
        else
            return 0;
    }
    if(n==2)//move down
    {
        for(int i=0;i<len;i++)
        {
            if(s[i]=='9')
            {
                pos=i;
                break;
            }
        }
        if(pos+3<9)
        {
            s[pos]=s[pos+3]^s[pos];
            s[pos+3]=s[pos]^s[pos+3];
            s[pos]=s[pos]^s[pos+3];

            return 1;
        }
        else
            return 0;
    }
    if(n==3)//move left
    {
        for(int i=0;i<len;i++)
        {
            if(s[i]=='9')
            {
                pos=i;
                break;
            }
        }
        if(pos-1>=0&&(pos/3==(pos-1)/3))
        {
            s[pos]=s[pos-1]^s[pos];
            s[pos-1]=s[pos]^s[pos-1];
            s[pos]=s[pos]^s[pos-1];

            return 1;
        }
        else
            return 0;
    }
    if(n==4)//move right
    {
        for(int i=0;i<len;i++)
        {
            if(s[i]=='9')
            {
                pos=i;
                break;
            }
        }
        if(pos+1<9&&(pos/3==(pos+1)/3))
        {
            s[pos]=s[pos+1]^s[pos];
            s[pos+1]=s[pos]^s[pos+1];
            s[pos]=s[pos]^s[pos+1];

            return 1;
        }
        else
            return 0;
    }

    return 0;
}

int bfs()
{
    while(!q.empty())
    {
        int ok=1;
        Note t=q.front();
        Note t2=q.front();
        q.pop();
        for(int i=0;i<8;i++)
        {
            if(t.s[i]-48==i+1&&t.s[8]=='9')
                   ;
            else
                {
                    ok=0;
                }
        }
 //       cout<<"判断是否可以结束:  "<<ok<<endl;

        if(ok)
        {
            return t.fa;
        }
 //           cout<<"还要继续搜索。。。"<<endl;
            t2=t;//cout<<"up up up\n";
            if(move4(t2.s,1))
            {
   //             cout<<"向上走。。。"<<endl;

                int k=cantor(t2.s);
                if(vis[k]==0)
                {
 //                   cout<<"行。。。"<<endl;
                    t2.dir='u';
                    t2.fa=t.num;
/*
                    cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
                    for(int i=0;i<9;i++)
                    {
                        cout<<t2.s[i]<<" ";
                        if((i+1)%3==0) cout<<endl;
                    }
                    cout<<t2.dir<<endl;

*/
                    t2.num=k;
                    note[k]=t2;
                    q.push(t2);
                    vis[k]=1;
                }
  //              else cout<<"不行。。。"<<endl;
            }
            t2=t; // cout<<"down down down\n";
            if(move4(t2.s,2))
            {
  //              cout<<"向下走。。。"<<endl;

                int k=cantor(t2.s);
                if(vis[k]!=1)
                {
                    t2.dir='d';
                    t2.fa=t.num;
  //                   cout<<"行。。。"<<endl;
/*
                    cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
                    for(int i=0;i<9;i++)
                    {
                        cout<<t2.s[i]<<" ";
                        if((i+1)%3==0) cout<<endl;
                    }
                    cout<<t2.dir<<endl;
*/
                    t2.num=k;
                    note[k]=t2;
                    q.push(t2);
                    vis[k]=1;
                }
  //              else cout<<"不行。。。"<<endl;
            }
            t2=t; // cout<<"left left left\n";
            if(move4(t2.s,3))
            {
  //              cout<<"向左走。。。"<<endl;

                int k=cantor(t2.s);
                if(vis[k]!=1)
                {
                    t2.dir='l';
                    t2.fa=t.num;
    //                 cout<<"行。。。"<<endl;
/*
                    cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
                    for(int i=0;i<9;i++)
                    {
                        cout<<t2.s[i]<<" ";
                        if((i+1)%3==0) cout<<endl;
                    }
                    cout<<t2.dir<<endl;

  */                  t2.num=k;
                    note[k]=t2;
                    q.push(t2);
                    vis[k]=1;
                }
   //            else cout<<"不行。。。"<<endl;
            }
            t2=t; //cout<<"right right right\n";
            if(move4(t2.s,4))
            {
   //             cout<<"向右走。。。"<<endl;

                int k=cantor(t2.s);
                if(vis[k]!=1)
                {
                    t2.dir='r';
                    t2.fa=t.num;
                    // cout<<"行。。。"<<endl;

/*
                    cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
                    for(int i=0;i<9;i++)
                    {
                        cout<<t2.s[i]<<" ";
                        if((i+1)%3==0) cout<<endl;
                    }
                    cout<<t2.dir<<endl;
*/
                    t2.num=k;
                    note[k]=t2;
                    q.push(t2);
                    vis[k]=1;
                }
  //              else cout<<"不行。。。"<<endl;
            }

    }

    return -999;
}


void print (int p)
{
    stack < char > stck;
    while ( note[p].dir!='Q' )
    {
        stck.push(note[p].dir);
        p = note[p].fa;
    }
    while ( !stck.empty() )
    {
        putchar ( stck.top() );
        stck.pop();
    }
    putchar ('\n');
}


int main()
{
    memset(vis,0,sizeof(vis));
    Note t;
    for(int i=0;i<9;i++)
    {
        char c;
        cin>>c;
        if(c=='x')
        {
            t.s[i]='9';
        }
        else t.s[i]=c;
    }

    int k=cantor(t.s);
    t.num=k;
    t.fa=-998;
    t.dir='Q';
    note[k]=t;

    q.push(t);

    int B=bfs();
    note[k].dir='Q';
    note[k].fa=-998;
    ///xxxxxx
    if(B==-999)  cout<<"unsolvable"<<endl;

  //  cout<<"B:  "<<B<<endl;

    print(0);

    return 0;
}