HDU-1043 java实现 单广+康托展开

       

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28459    Accepted Submission(s): 7557
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

这是一道经典的搜索题，在网上一搜一大把的解析……然鹅，一开始我只是想做道简单的BFS练练手先……没想到这道题如此“经典”，直接刚了一整天……下面开始我的总结。（新手向）

在做这道题之前，首先你得有预备知识，才能做出这道题。（康托展开/双向BFS/A*/逆序数奇偶性，怎么着你也得会一种，康托展开必备）

很明显这是一道搜索题。那么采用深搜还是宽搜呢？这个很显然不能用深搜。交换两个数，深搜可以深到天际，量太大了。而宽搜可以更快地遍历到所有的状态。我们的目标就是把所有的362880种状态都找出来，这点宽搜的缺点反而变成了优点。

好，我们得出结论，用宽搜。然后你会萌萌地发现，终止条件是啥啊？？呵呵，这时候我们悄悄看一眼DISCUSS，发现一个词“康托展开”。事实上，这道题不是用宽搜搜出答案来的。我们只是用宽搜去打表。我们会发现，所有的状态一共只有9！个，其实是有限的。而每个状态到达目标状态的方式都是唯一的。（这么说不太准确，其实方式可能有多种，但是对于某一种特点的解题策略而言，得出的结论应该是一样的。）所以说，思路来了：如果我们能求出那9！种状态各自的方法，那么只要把它存起来，输入哪种状态就输出答案就好了，绝对省时。

当然，萌新全凭自己写出代码还是不容易的。先看懂别人写的才是王道。

这里贴上代码，暂时只有一种，单向BFS+康托打表，我认为还是比较好理解的：

import java.util.Scanner;  
import java.util.LinkedList;
public class Main {
    static int[] fac={1,1,2,6,24,120,720,5040,40320,362880};
    static NodeWay[] nn=new NodeWay[363000];
    static int[][] dir={{0,1},{1,0},{-1,0},{0,-1}};
    
    
	public static void main(String[] args) {
		// TODO 自动生成的方法存根
       Scanner cin=new Scanner(System.in);
       for(int i=0;inumber[j])count++;
			}
			result+=count*fac[n-i-1];
		}
		return result+1;
	}
    
	static void bfs()    //从123456780的末状态开始往之前的状态搜，搜出一种不同的就给他存起来
	{
		Node start=new Node();
		start.CantorValue=46234;
		for(int i=0;i<9;i++)
		{
			start.map[i]=(i+1)%9;
		}
		start.loc=8;
		nn[46234].Father=0;
		int row;
		int col;
		LinkedList queue=new LinkedList();
		queue.add(start);
		while(!queue.isEmpty())
		{
			Node node=queue.poll();
			for(int i=0;i<4;i++)
			{
				row=node.loc/3+dir[i][0];
				col=node.loc%3+dir[i][1];
				if(row<0||col<0||row>=3||col>=3)continue;
				Node work=new Node();
				work.CantorValue=node.CantorValue;
				work.loc=node.loc;
				for(int k=0;k<9;k++){
					work.map[k]=node.map[k];
				}
				work.loc=row*3+col;
				int temp=0;
				temp=work.map[node.loc];
				work.map[node.loc]=work.map[work.loc];
				work.map[work.loc]=temp;
				work.CantorValue=cantor(work.map,9);
				if(nn[work.CantorValue].Father==-1)
				{
					nn[work.CantorValue].Father=node.CantorValue;
					if(i==0)nn[work.CantorValue].Way='l';
					if(i==1)nn[work.CantorValue].Way='u';
					if(i==2)nn[work.CantorValue].Way='d';
					if(i==3)nn[work.CantorValue].Way='r';
					queue.offer(work);
				}
			}
		}
	}
}

class NodeWay
{
	char Way;
	int Father;
	NodeWay(){}
}

class Node
{
	int CantorValue;           //康托值
	int[] map=new int[9];      //当前状态。用一维数组表示。
	int loc;     //标记出x的位置，方便操作
	Node(){}
}