hdu 1043 八数码问题

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10778    Accepted Submission(s):
2873
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)

 

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 康托展开优化，代码自己写的过了，做EIGHTII的时候，发现别人bfs（）很简单，而且map[4][2]写的看不懂。

 

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<string>
 #include<queue>
 using namespace std;

 struct node
 {
     bool flag;
     char str;
     int father;
 }hash[];
 struct st
 {
     char t[];
 };
 queue<st>Q;
 int ans[]={};

 int ktzk(char *c)
 {
     int i,j,k,sum=;
     for(i=;i<=;i++)
     {
         k=;
         for(j=i+;j<=;j++)
             if(c[i]>c[j])
                 k++;
         sum=sum+k*ans[-i];
     }
     return sum;
 }
 void bfs()
 {
     int i,k,num,val;
     struct st cur,t;
     k=ktzk("");
     hash[k].flag=true;
     hash[k].str='\0';
     hash[k].father=k;

     strcpy(t.t,"");
     Q.push(t);

     while(!Q.empty())
     {
         cur=Q.front();
         Q.pop();
         val=ktzk(cur.t);
         for(i=;i<=;i++)
         {
             if(cur.t[i]=='')
             {
                 k=i;
                 break;
             }
         }
         if(k!= && k!= && k!=)//rigth
         {
             t=cur;
             swap(t.t[k],t.t[k+]);
             num=ktzk(t.t);
             if(hash[num].flag==false)
             {
                 hash[num].flag=true;
                 hash[num].str='r';
                 hash[num].father=val;
                 Q.push(t);
             }
         }
         if(k!= && k!= && k!=)//left
         {
             t=cur;
             swap(t.t[k],t.t[k-]);
             num=ktzk(t.t);
             if(hash[num].flag==false)
             {
                 hash[num].flag=true;
                 hash[num].str='l';
                 hash[num].father=val;
                 Q.push(t);
             }
         }
         if(k>=)//u
         {
             t=cur;
             swap(t.t[k],t.t[k-]);
             num=ktzk(t.t);
             if(hash[num].flag==false)
             {
                 hash[num].flag=true;
                 hash[num].str='u';
                 hash[num].father=val;
                 Q.push(t);
             }
         }
         if(k<=)//D
         {
             t=cur;
             swap(t.t[k],t.t[k+]);
             num=ktzk(t.t);
             if(hash[num].flag==false)
             {
                 hash[num].flag=true;
                 hash[num].str='d';
                 hash[num].father=val;
                 Q.push(t);
             }
         }
     }
 }
 void prepare()
 {
     int i;
     for(i=;i<=;i++)
     {
         hash[i].flag=false;
     }
     for(i=;i<=;i++) ans[i]=ans[i-]*i;
     bfs();
 }
 int main()
 {
     prepare();
     char a[],b[],c[];
     bool tom[];
     int i,j,k,cur;
     while(gets(a))
     {
         k=strlen(a);
         b[]='\0';
         for(i=,j=;i<k;i++)
         {
             if(a[i]=='x' || (a[i]>=''&&a[i]<=''))
             {
                 if(a[i]=='x')
                     b[j]='';
                 else b[j]=a[i];
                 j++;
             }
         }
         memset(tom,false,sizeof(tom));
         for(i=;i<=;i++)
         {
             tom[b[i]-'']=true;
         }
         for(i=;i<=;i++)
         {
             if(tom[i]==false)
                 break;
         }
         if(i<=){printf("unsolvable\n");continue;}
         cur=ktzk(b);
         if(hash[cur].flag==false)
         {
             printf("unsolvable\n");
             continue;
         }
         k=;
         while(hash[cur].father!=cur)
         {
             c[k]=hash[cur].str;
             k++;
             cur=hash[cur].father;
         }
         for(i=;i<=k-;i++)
         {
             if(c[i]=='u')printf("d");
             if(c[i]=='d')printf("u");
             if(c[i]=='l')printf("r");
             if(c[i]=='r')printf("l");
         }
         printf("\n");
     }
     return ;
 }