HDU 1043 Eight【A*算法】 【康托展开】【八码问题】 【哈希】
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31183 Accepted Submission(s): 8176
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)
Recommend
JGShining
关于八码问题的介绍：
https://blog.csdn.net/huangxy10/article/details/8034285
关于A*算法的介绍：
http://www.cppblog.com/mythit/archive/2009/04/19/80492.aspx
在做此八码问题之前，先了解八码问题的一个可解问题：
（此段话摘自博客：https://blog.csdn.net/huangxy10/article/details/8034285）
八数码问题的一个状态实际上是0~9的一个排列，对于任意给定的初始状态和目标，不一定有解，也就是说从初始状态不一定能到达目标状态。因为排列有奇排列和偶排列两类，从奇排列不能转化成偶排列或相反。
如果一个数字0~8的随机排列871526340，用F(X)表示数字X前面比它小的数的个数，全部数字的F(X)之和为Y=∑(F(X))，如果Y为奇数则称原数字的排列是奇排列，如果Y为偶数则称原数字的排列是偶排列。
例如871526340这个排列的
Y=0+0+0+1+1+3+2+3+0=10
10是偶数，所以他偶排列。871625340
Y=0+0+0+1+1+2+2+3+0=9
9是奇数，所以他奇排列。
所以说明：当原排列是奇排列时无解。
如何用康托展开来计算哈希呢？
(此段话摘自于博客：https://blog.csdn.net/iceiceicpc/article/details/52119994）
首先这题必须要用hash和康托展开来压缩空间，康托展开的原理我用一个例子来讲解：
对于这样一个排列：14032，他在排列中排第几呢？公式是：1*4!+3*3!+0*2!+1*1!+0*0!
求法是这样的，首先看第一个数字1，在数字1前面（不是这个排列的前面，而是这个排列所出现的数字中，如14032中只有一个0比它小）小于它的数只有一个0，对于它后面的数可以有4！种（4的全排列），所以对于1就是1*4！，同理对于4，0~4比它小的有0,1,2,3，但是1在前面已经用过了，所以只有3个，对于后面就是3！，所以对于4就是3*3！，同理往后推。
但是为什么要用hash呢，其实就是为了判重，因为如果你用普通的方法去判重，那么对于3*3方块，每一个数字都有9种可能0~8，况且你用普通bfs和dfs的vis数组根本无法判这题状态是否重复。但是hash可以很好的解决这个问题，因为hash可以把3*3这些数字全部排列的个数控制在0~9！上，对于每一个转态就有一个唯一的hash值，此时只要开一个一维的vis[hash最大值]数组就
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