hud 1043 Eight

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14009    Accepted Submission(s): 3965
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, 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
   
   
   
   

 

八数码问题。

BFS+康拓展开

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#define N 400000
using namespace std;

struct node
{
    int pos,val;
    char a[9];
    string path;
}cur,now;

int f[15]= {40320,5040,720,120,24,6,2,1,1}; //康拖展开判重

int dx[4]={-1,1,0,0};//上 下 左 右
int dy[4]={0,0,-1,1};
char dir[5]="durl";

string anspath[N],ans;

int vis[N];
char c;

queue<node>q;

int kanto(char *ss)
{
    int ans=0;
    int num;

    for(int i=0;i<9;i++)
    {
        num=0;
        for(int j=i+1;j<9;j++)
        {
            if(ss[i]>ss[j]) num++;
        }
        ans+=num*f[i];
    }
    return ans+1;
}


void bfs()
{
    int x,y,t,t1,tempv;
    char temps[9];

    memset(vis,0,sizeof vis);
     while(!q.empty())q.pop();

    for(int i=0;i<8;i++)
    cur.a[i]=i+1-0+'0';


    cur.a[8]='0';
    cur.pos=9;
    cur.path="";
    cur.val=kanto(cur.a);
    vis[cur.val]=1;
    anspath[cur.val]=cur.path;

    q.push(cur);

    while(!q.empty())
    {
        now=q.front();
        t=now.pos;
        y=t%3;
        if(y==0) y=3;
        x=(t-y)/3+1;

        q.pop();
        int nx,ny;

        for(int i=0;i<4;i++)
        {
            nx=x+dx[i];
            ny=y+dy[i];//眼睛都看瞎了，才看到写成n=y=dy[i];(⊙o⊙)…
            t1=(nx-1)*3+ny;
            memcpy(temps,now.a,sizeof(temps));
            temps[t-1]=now.a[t1-1]; //模拟移动
            temps[t1-1]='0';

            tempv=kanto(temps);

            if(nx>=1 && nx<=3 && ny>=1 && ny<=3 && !vis[tempv])
            {
                vis[tempv]=1;
                cur.pos=t1;
                cur.path=now.path+dir[i];
                memcpy(cur.a,temps,sizeof(temps));
                cur.val=tempv;
                q.push(cur);
                anspath[cur.val]=cur.path;
            }
        }
    }

}

int main()
{
//    f[0]=1;
//    for(int i=1;i<9;i++)
//    f[i]=f[i-1]*i;

    bfs();

    while(cin>>c)
    {
        if(c=='x')
        {
            cur.pos=1;
            cur.a[0]='0';
        }
        else
        cur.a[0]=c;


        for(int i=1;i<9;i++)
        {
            cin>>c;
            if(c=='x')
            {
                cur.pos=i+1;
                cur.a[i]='0';
            }
            else
            cur.a[i]=c;
        }

        cur.val=kanto(cur.a);
        int flag;

        if(vis[cur.val])
        {
            flag=anspath[cur.val].length();
            for(int i=flag-1;i>=0;i--)
             printf("%c",anspath[cur.val][i]);

        }
        else
        printf("unsolvable");
        cout<<endl;

    }

    return 0;
}