hdu 1034 & poj 1077 Eight 传说中的八数码问题。真是一道神题，A*算法+康托展开

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13506    Accepted Submission(s): 3855
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
   
   
   
   

 

不得不说，，人这一生，要不是不刷这题，简直不完整啊。附上大量资料.。要想做这题，，得好好准备啊！！

A*算法入门：http://www.policyalmanac.org/games/Chine%20Translation%20-%20For%20beginners.html

本题详细解析请看：http://blog.csdn.net/acm_cxlove/article/details/7745323

康托展开请看：康托展开

八数码八大境界：八数码的八境界

哎。做的时候可是相当痛苦啊。

下面是我的代码：

#include <cstdio>
#include <queue>
#include <cstring>
#include<algorithm>
#include <string>
#define MAX 3

using namespace std ;

struct Node{
	int map[MAX][MAX] ,hash;
	int f,g,h;
	int x,y;
	/*bool operator<(const Node &n)const
	{
		return f>n.f ;
	}*/
	//下面比上面更快 
	bool operator<(const Node n1)const{     //优先队列第一关键字为h,第二关键字为g
		return h!=n1.h?h>n1.h:g>n1.g;
	}
	bool check()
	{
		if(x<0||y<0 || x>=MAX||y>=MAX)
		{
			return false ;
		}
		return true ;
	}
};
const int HASH[9]={1,1,2,6,24,120,720,5040,40320};   //HASH的权值  
const int dir[4][2]={1,0,-1,0,0,-1,0,1} ;
int visited[400000] ;
int pre[400000] ;
int des = 322560 ;
int getHash(Node n)
{
	int oth[MAX*4] , k = 0;
	for(int i = 0 ; i < MAX ; ++i)
	{ 
		for(int j = 0 ; j < MAX ; ++j)
		{
			oth[k++] = n.map[i][j] ;
		}
	}
	int result = 0 ;
	for(int i = 0 ; i < 9 ; ++i)
	{
		int count = 0 ; 
		for(int j = 0 ; j < i ; ++j)
		{
			if(oth[i]<oth[j])
			{
				count++;
			}
		}
		result += count*HASH[i] ;
	}
	return result ;
}

int getH(Node n)
{
	int result = 0 ;
	for(int i = 0 ; i < MAX ; ++i)
	{
		for(int j = 0 ; j < MAX ; ++j)
		{
			if(n.map[i][j])
			{
				int x = (n.map[i][j]-1)/3 , y = (n.map[i][j]-1)%3 ;
				result += abs(x-i)+abs(y-j) ;
			}
		}
	}
	return result ;
}

bool judge(Node n)
{
	int oth[MAX*4] , k = 0;
	for(int i = 0 ; i < MAX ; ++i)
	{ 
		for(int j = 0 ; j < MAX ; ++j)
		{
			oth[k++] = n.map[i][j] ;
		}
	}
	int result = 0 ;
	for(int i = 0 ; i < 9 ; ++i)
	{
		for(int j = i+1 ; j < 9 ; ++j)
		{
			if(oth[i]&&oth[j]&&oth[i]>oth[j])
			{
				++result;
			}
		}
	}
	return !(result&1) ;
}

void AStar(Node start)
{
	priority_queue<Node> p;
	p.push(start);
	while(!p.empty())
	{
		Node n = p.top();
		p.pop();
		for(int i = 0 ; i < 4 ; ++i)
		{
			Node next = n;
			next.x += dir[i][0];
			next.y += dir[i][1];
			if(!next.check())
			{
				continue ;
			}
			swap(next.map[next.x][next.y],next.map[n.x][n.y]) ;
			next.hash = getHash(next) ;
			if(visited[next.hash] == -1)
			{
				next.h = getH(next) ;
				next.g++ ;
				next.f = next.g+next.h;
				pre[next.hash] = n.hash ;
				p.push(next) ;
				visited[next.hash] = i ;	//i代表方向 
			}
			if(next.hash == des)
			{
				return ;
			}
		}
	}
}

void print()
{
	int next = des ;
	string ans;
	ans.clear() ;
	while(pre[next]!=-1)
	{
		switch(visited[next])
		{
			case 0 : ans += 'd' ; break ;
			case 1 : ans += 'u' ; break ;
			case 2 : ans += 'l' ; break ;
			case 3 : ans += 'r' ; break ;
			default : break ; 
		}
		next = pre[next] ;
	}
	int len = ans.size() ;
	for(int i = len-1 ; i >=0 ; --i)
	{
		putchar(ans[i]) ;
	}
	puts("");
}

int main()
{
	char str[100] ;
	while(gets(str) != NULL)
	{
		Node t ;
		memset(visited,-1,sizeof(visited)) ;
		memset(pre,-1,sizeof(pre)) ;
		int k = 0 ,i = 0;
		while(str[k] != '\0')
		{
			if(str[k]>'0'&&str[k]<='9')
			{
				t.map[i/3][i%3] = str[k]-'0' ;
				++i ;
			}
			else if(str[k] == 'x')
			{
				t.x = i/3 ;
				t.y = i%3 ;
				t.map[i/3][i%3] = 0 ;
				++i ;
			}
			++k ;
		}
		t.hash=getHash(t);
		visited[t.hash] = -2 ;
		t.g = 0 ;
		t.h = getH(t);
		t.f = t.g+t.h;
		if(!judge(t))
		{
			printf("unsolvable\n"); 
			continue ;
		}
		if(t.hash == des)
		{
			puts(""); 
			continue ;
		}
		AStar(t) ;
		print();
	}
	return 0;
}