HDU 1043 Eight（bfs+康托展开）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15216    Accepted Submission(s): 4235
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

/*

对于一个9的序列，变成3*3宫格,x表示可以移动的，求这样使他回到有序
123
456
780

思路：康托hash，然后path就录当前状态是由哪一个状态变来，
                然后di[][]表示当前状态由前一个状态由哪一个方向变来

*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;
#define bug printf("hi\n");
using namespace std;

#define INF 0x3f3f3f3f
#define N  400000

int fac[10];
int di[N],pre[N];
int vis[N];

struct stud{
 int a[10];
}f[N];

int step[4][2]={1,0,-1,0,0,1,0,-1};
char c[]="udlr";       //因为要逆着走，所以方向相反


int start;

void inint()
{
    int i,j;
    fac[0]=1;
    for(i=1;i<9;i++)
        fac[i]=fac[i-1]*i;
}

int cantor(int a[])
{
    int temp,i,j;
    temp=0;
    int cnt;
    for(i=0;i<9;i++)
    {
       cnt=0;
       for(j=i+1;j<9;j++)
            if(a[i]>a[j]) cnt++;
       temp+=cnt*fac[9-i-1];
    }
  return temp;
}

void bfs()
{
    int i,j;
    int head,tail;
    stud cur,next;

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

    start=cantor(cur.a);

    memset(vis,0,sizeof(vis));
    vis[start]=1;

    head=tail=0;
    f[tail++]=cur;
    while(head<tail)   //这里最好不要用queue，不然会超时
    {
        cur=f[head];
        head++;
        for(i=0;i<9;i++)
            if(cur.a[i]==0) break;

        int x=i/3;
        int y=i%3;

        for(i=0;i<4;i++)
        {
            int xx=x+step[i][0];
            int yy=y+step[i][1];
            int id=cantor(cur.a);
            if(xx<0||xx>=3||yy<0||yy>=3) continue;
            next=cur;
            swap(next.a[x*3+y],next.a[xx*3+yy]);
            int temp=cantor(next.a);
            if(!vis[temp])
            {
                vis[temp]=1;
                pre[temp]=id;
                di[temp]=i;
                f[tail++]=next;
            }
            swap(next.a[x*3+y],next.a[xx*3+yy]);
        }
    }
}

int main()
{
    inint();
    int i,j;
    bfs();

    int a[10];
    char s[10];
    while(~scanf("%1s",s))
    {
        if(s[0]=='x') a[0]=0;
        else a[0]=s[0]-'0';

        for(i=1;i<9;i++)
        {
            scanf("%1s",s);
            if(s[0]=='x') a[i]=0;
            else a[i]=s[0]-'0';
        }

        int temp=cantor(a);
        if(!vis[temp])
            printf("unsolvable\n");
        else
        {
           while(temp!=start)
           {
               printf("%c",c[di[temp]]);
               temp=pre[temp];
           }
           printf("\n");
        }
    }
     return 0;
}