1043

Posted 2022-11-26 tookkke

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18728    Accepted Submission(s): 5058
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，询问很多次，但终态是相同的，所以可以一开始逆向BFS求出所有答案(其实就是打表)。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

const int fact[]=1,1,2,6,24,120,720,5040,40320,362880;
const int mv[4]=-3,-1,3,1;
const char *const dir="drul";

char hehe[362880];
int path[362880];

struct theStep
	int kkke[9];
	int x;
	int cnt;
	bool operator==(const theStep &a)const
	
		for(int i=0;i<9;i++)
		    if(kkke[i]!=a.kkke[i])
		        return false;
		return true;
	
	bool moveto(theStep &a,int b)
	
		if(x+b<0||x+b>=9||(abs(b)==1&&x/3!=(x+b)/3))return false;
		for(int i=0;i<9;i++)a.kkke[i]=kkke[i];
		swap(a.kkke[x],a.kkke[x+b]);
		a.x=x+b;
		a.cnt=cnt+1;
		return true;
	
	int getValue()
	
		int value=0;
		for(int i=1;i<9;i++)
			for(int j=0;j<i;j++)
			    if(kkke[i]>kkke[j])
			        value+=fact[i];
		return value;
	
S,T,nows,nexts;
int SValue;

queue<theStep>q;

void init()

	for(int i=0;i<8;i++)S.kkke[i]=i+1;
	S.kkke[8]=0;
	S.cnt=0;
	S.x=8;
	SValue=S.getValue();
	q.push(S);
	hehe[SValue]='l';
	while(!q.empty())
	
		nows=q.front();q.pop();
		int nowValue=nows.getValue();
		for(int i=0;i<4;i++)
		
			if(nows.moveto(nexts,mv[i]))
			
				int nextValue=nexts.getValue();
				if(hehe[nextValue])continue;
				q.push(nexts);
				hehe[nextValue]=dir[i];
				path[nextValue]=nowValue;
			
		
	


int main()

	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	init();
	int a[10];
	while(scanf("%s",a)==1)
	
		if(a[0]=='x')T.kkke[0]=0;
		else T.kkke[0]=a[0]-'0';
		for(int i=1;i<9;i++)
		
			scanf("%s",a);
			if(a[0]=='x')T.kkke[i]=0;
		    else T.kkke[i]=a[0]-'0';
		
		if(S==T)printf("lr\\n");
		else
		
			int nowv=T.getValue();
			if(hehe[nowv])
			
				do
			        putchar(hehe[nowv]);
			        nowv=path[nowv];
			    while(nowv!=SValue);
		        putchar('\\n');
			
			else printf("unsolvable\\n");
		    
	    
	
	return 0;