infinity

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http://acm.pku.edu.cn/JudgeOnline/problem?id=1077

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

South Central USA 1998


八数码 经典的BFS啊, 我就直接用的单向的BFS,也过了,双向的应该会快很多。
判重的hash函数看的discuss里的,用逆序数,9!种情况,一一对应,不会有重复的。


 

Source Code
Problem: 1077
User: lovecanon
Memory: 9572K
Time: 125MS
Language: GCC
Result: Accepted
  • Source Code
  • 
        
    #include<stdio.h>
    #include
    <string.h>
    #include
    <stdlib.h>
    struct node
    {
        
    int state[3][3];
        
    int pre;
        
    int dir;
    }queue[
    362881];
    int hash[362881];
    int step[362881];
    int a[3][3];

    int fac(int i)
    {
        
    switch(i)
        {
            
    case 0return 1;
            
    case 1return 1;
            
    case 2return 2;
            
    case 3return 6;
            
    case 4return 24;
            
    case 5return 120;
            
    case 6return 720;
            
    case 7return 5040;
            
    case 8return 40320;
        }
        
    return 0;
    }

    int HASH()
    {
        
    int i,j,k=0,b[9],ret=0,num=0;
        
    for(i=0;i<3;i++)
            
    for(j=0;j<3;j++)
                b[k
    ++]=a[i][j];
        
    for(i=0;i<9;i++)
        {
            num
    =0;
            
    for(j=0;j<i;j++)
                
    if(b[j]>b[i])  num++;
            ret
    +=fac(i)*num;
        }
        
    return ret;
    }

    void output(int len)
    {
        
    int i;
        
    for(i=len;i>=0;i--
        {
            
    if(step[i]==1) printf("l");
            
    if(step[i]==2) printf("r");
            
    if(step[i]==3) printf("u");
            
    if(step[i]==4) printf("d");
        }
        printf(
    "\n");
    }

    int main()
    {
        
    char s[10];
        
    int i,j,rear=0,front=0,tag=0;

        rear
    ++;
        
    for(i=0;i<3;i++)
            
    for(j=0;j<3;j++)
            {
                scanf(
    "%s",s);
                
    if(s[0]=='x')  s[0]='9';
                queue[rear].state[i][j]
    =s[0]-'0';
            }
        queue[rear].pre
    =0;queue[rear].dir=0;
        
    for(i=0;i<3;i++)
            
    for(j=0;j<3;j++)
                a[i][j]
    =queue[rear].state[i][j];

        hash[HASH()]
    =1;
        
    while(front<rear)
        {
            
    int e,f,tmp,cntdir,len;
            front
    ++;
            
    for(i=0;i<3;i++)
                
    for(j=0;j<3;j++)
                {
                    a[i][j]
    =queue[front].state[i][j];
                    
    if(a[i][j]==9) {e=i;f=j;}
                }
            
    if(f-1>=0)
            {
                cntdir
    =1;
                tmp
    =a[e][f];
                a[e][f]
    =a[e][f-1];
                a[e][f
    -1]=tmp;
                tmp
    =HASH();
                
    if(tmp==0
                {
                    
    int t=front;
                    len
    =0;
                    step[len
    ++]=cntdir;
                    
    while(queue[t].pre)  
                    {
                        step[len
    ++]=queue[t].dir;
                        t
    =queue[t].pre;
                    }
                    output(len);
                    
    return 0;
                }
                
    if(!hash[tmp]) 
                {
                    rear
    ++;
                    
    for(i=0;i<3;i++)
                        
    for(j=0;j<3;j++)
                            queue[rear].state[i][j]
    =a[i][j];
                    queue[rear].dir
    =cntdir;queue[rear].pre=front;
                    hash[tmp]
    =1;

                }
                tmp
    =a[e][f];
                a[e][f]
    =a[e][f-1];
                a[e][f
    -1]=tmp;
            }
            
    if(f+1<3)
            {
                cntdir
    =2;
                tmp
    =a[e][f];
                a[e][f]
    =a[e][f+1];
                a[e][f
    +1]=tmp;
                tmp
    =HASH();
                
    if(tmp==0
                {
                    
    int t=front;
                    len
    =0;
                    step[len
    ++]=cntdir;
                    
    while(queue[t].pre)  
                    {
                        step[len
    ++]=queue[t].dir;
                        t
    =queue[t].pre;
                    }
                    output(len);
                    
    return 0;
                }
                
    if(!hash[tmp]) 
                {
                    rear
    ++;
                    
    for(i=0;i<3;i++)
                        
    for(j=0;j<3;j++)
                            queue[rear].state[i][j]
    =a[i][j];
                    queue[rear].dir
    =cntdir;queue[rear].pre=front;
                    hash[tmp]
    =1;
                }
                tmp
    =a[e][f];
                a[e][f]
    =a[e][f+1];
                a[e][f
    +1]=tmp;
            }
            
    if(e-1>=0)
            {
                cntdir
    =3;
                tmp
    =a[e][f];
                a[e][f]
    =a[e-1][f];
                a[e
    -1][f]=tmp;
                tmp
    =HASH();
                
    if(tmp==0
                {
                    
    int t=front;
                    len
    =0;
                    step[len
    ++]=cntdir;
                    
    while(queue[t].pre)  
                    {
                        step[len
    ++]=queue[t].dir;
                        t
    =queue[t].pre;
                    }
                    output(len);
                    
    return 0;
                }
                
    if(!hash[tmp]) 
                {
                    rear
    ++;
                    
    for(i=0;i<3;i++)
                        
    for(j=0;j<3;j++)
                            queue[rear].state[i][j]
    =a[i][j];
                    queue[rear].dir
    =cntdir;queue[rear].pre=front;
                    hash[tmp]
    =1;

                }
                tmp
    =a[e][f];
                a[e][f]
    =a[e-1][f];
                a[e
    -1][f]=tmp;
            }
            
    if(e+1<3)
            {
                cntdir
    =4;
                tmp
    =a[e+1][f];
                a[e
    +1][f]=a[e][f];
                a[e][f]
    =tmp;
                tmp
    =HASH();
                
    if(tmp==0
                {
                    
    int t=front;
                    len
    =0;
                    step[len
    ++]=cntdir;
                    
    while(queue[t].pre)  
                    {
                        step[len
    ++]=queue[t].dir;
                        t
    =queue[t].pre;
                    }
                    output(len);
                    
    return 0;
                }
                
    if(!hash[tmp]) 
                {
                    rear
    ++;
                    
    for(i=0;i<3;i++)
                        
    for(j=0;j<3;j++)
                            queue[rear].state[i][j]
    =a[i][j];
                    queue[rear].dir
    =cntdir;queue[rear].pre=front;
                    hash[tmp]
    =1;

                }
                tmp
    =a[e+1][f];
                a[e
    +1][f]=a[e][f];
                a[e][f]
    =tmp;
            }
        }
        printf(
    "unsolvable\n");
        
    return 0;
    }

posted on 2008-09-20 04:18 infinity 阅读(1866) 评论(0)  编辑 收藏 引用 所属分类: acm

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