Eight

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8536    Accepted Submission(s): 2321
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
  

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 362881
struct Node
{
    int state[3][3];
    int pre,dir;
    void printState()
    {
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                printf("%d ", state[i][j]);
            }
            printf("/n");
        }
        printf("/n");
    }
}queue[MAX];
int cantor[9]={1,1,2,6,24,120,720,5040,40320};
int rear, front, step[MAX], len;
bool visited[MAX];
int hash(int a[3][3])
{
    int b[9],k=0,result=0;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            b[k++]=a[i][j];
        }
    }
    for(int i=0;i<9;i++)
    {
        int num=0;
        for(int j=0;j<i;j++)
        {
            if(b[j]>b[i])
            {
                num++;
            }
        }
        result+=cantor[i]*num;
    }
    return result;
}
void input()
{
    char s[10];
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            scanf("%s",s);
            if(s[0]=='x')  s[0]='9';
            queue[0].state[i][j]=s[0]-'0';
        }
    }
}
void makeResult()
{
    len=0;
    while(queue[front].pre!=-1)
    {
        step[len++]=queue[front].dir;
        front=queue[front].pre;
    }
}
void output()
{
    for(int i=len-1;i>=0;i--)
    {
        switch(step[i])
        {
            case 0: printf("l"); break;
            case 1: printf("u"); break;
            case 2: printf("r"); break;
            case 3: printf("d"); break;
        }
    }
    printf("/n");
}
bool complete()
{
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            if(i*3+j+1!=queue[front].state[i][j])
            {
                return false;
            }
        }
    }
    return true;
}
bool bfs()
{
    queue[0].pre=-1;
    queue[0].dir=-1;
    rear=0,front=1;
    visited[hash(queue[0].state)]=true;
    if(complete()) return true;
    while(rear<front)
    {
        int nx,ny;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                if(queue[rear].state[i][j]==9)
                {
                    nx=i,ny=j;
                }
            }
        }
        if(ny>0)
        {
            for(int i=0;i<3;i++)
            {
                for(int j=0;j<3;j++)
                {
                    queue[front].state[i][j]=queue[rear].state[i][j];
                }
            }
            queue[front].state[nx][ny]^=queue[front].state[nx][ny-1];
            queue[front].state[nx][ny-1]^=queue[front].state[nx][ny];
            queue[front].state[nx][ny]^=queue[front].state[nx][ny-1];
            int temp=hash(queue[front].state);
            if(!visited[temp])
            {
                visited[temp]=true;
                queue[front].pre=rear;
                queue[front].dir=0;
                if(complete())
                {
                    return true;
                }
                front++;
            }
        }
        if(nx>0)
        {
            for(int i=0;i<3;i++)
            {
                for(int j=0;j<3;j++)
                {
                    queue[front].state[i][j]=queue[rear].state[i][j];
                }
            }
            queue[front].state[nx][ny]^=queue[front].state[nx-1][ny];
            queue[front].state[nx-1][ny]^=queue[front].state[nx][ny];
            queue[front].state[nx][ny]^=queue[front].state[nx-1][ny];
            int temp=hash(queue[front].state);
            if(!visited[temp])
            {
                visited[temp]=true;
                queue[front].pre=rear;
                queue[front].dir=1;
                if(complete())
                {
                    return true;
                }
                front++;
            }
        }
        if(ny<2)
        {
            for(int i=0;i<3;i++)
            {
                for(int j=0;j<3;j++)
                {
                    queue[front].state[i][j]=queue[rear].state[i][j];
                }
            }
            queue[front].state[nx][ny]^=queue[front].state[nx][ny+1];
            queue[front].state[nx][ny+1]^=queue[front].state[nx][ny];
            queue[front].state[nx][ny]^=queue[front].state[nx][ny+1];
            int temp=hash(queue[front].state);
            if(!visited[temp])
            {
                visited[temp]=true;
                queue[front].pre=rear;
                queue[front].dir=2;
                if(complete())
                {
                    return true;
                }
                front++;
            }
        }
        if(nx<2)
        {
            for(int i=0;i<3;i++)
            {
                for(int j=0;j<3;j++)
                {
                    queue[front].state[i][j]=queue[rear].state[i][j];
                }
            }
            queue[front].state[nx][ny]^=queue[front].state[nx+1][ny];
            queue[front].state[nx+1][ny]^=queue[front].state[nx][ny];
            queue[front].state[nx][ny]^=queue[front].state[nx+1][ny];
            int temp=hash(queue[front].state);
            if(!visited[temp])
            {
                visited[temp]=true;
                queue[front].pre=rear;
                queue[front].dir=3;
                if(complete())
                {
                    return true;
                }
                front++;
            }
        }
        rear++;
    }
    return false;
}
int main()
{
    input();
    if(bfs())
    {
        makeResult();
        output();
    }
    else
    {
        printf("unsolvable/n");
    }
    return 0;
}