POJ 1077 Eight（康托展开+BFS）

Posted 2020-08-04 Blackops

Eight

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30176		Accepted: 13119		Special Judge

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr

Source

South Central USA 1998

 

题目链接：POJ 1077

主要就是用康托展开来映射判重的问题，info::val就是康托展开hash值，info::step就是积累的状态。另外感觉这题剧毒，自己本来用int vis[]和char his[]想最后回溯记录答案从而代替速度比较慢的string，结果居然超时……TLE一晚上，要不是看了大牛的博客估计要一直T在这个坑点上。还有不知道为什么string的加号重载在C++编译器里会CE，换G++才过。

什么是康托展开？——康托展开介绍文章

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=362880+20;
int fact[10]={1,1,2,6,24,120,720,5040,40320,362880};
int direct[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char MOVE[5]="udlr";
struct info
{
	int s[9];
	int indx;
	string step;
	int val;
};
info S,E;
int T;
int vis[N];
string ans;
int calcantor(int s[])
{
	int r=0;
	for (int i=0; i<9; ++i)
	{
		int k=0;
		for (int j=i+1; j<9; ++j)
		{
			if(s[j]<s[i])
				++k;
		}
		r=r+k*fact[8-i];
	}
	return r;
}
bool check(const int &x,const int &y)
{
	return (x>=0&&x<3&&y>=0&&y<3);
}
bool bfs()
{
	CLR(vis,0);
	queue<info>Q;
	Q.push(S);
	vis[S.val]=1;
	info now,v;
	while (!Q.empty())
	{
		now=Q.front();
		Q.pop();
		if(now.val==T)
		{
			ans=now.step;
			return true;
		}
		for (int i=0; i<4; ++i)
		{
			int x=now.indx/3;
			int y=now.indx%3;
			x+=direct[i][0];
			y+=direct[i][1];
			if(check(x,y))
			{
				v=now;
				v.indx=x*3+y;
				v.s[now.indx]=v.s[v.indx];
				v.s[v.indx]=0;		
				v.val=calcantor(v.s);
				if(!vis[v.val])
				{				
					vis[v.val]=1;
					v.step=now.step+MOVE[i];
					if(v.val==T)
					{
						ans=v.step;
						return true;
					}
					Q.push(v);
				}
			}
		}
	}
	return false;
}
int main(void)
{
	char temp;
	int i;
	T=46233;
	while (cin>>temp)
	{
		if(temp==‘x‘)
		{
			S.s[0]=0;
			S.indx=0;
		}
		else
			S.s[0]=temp-‘0‘;
		for (i=1; i<9; ++i)
		{
			cin>>temp;
			if(temp==‘x‘)
			{
				S.s[i]=0;
				S.indx=i;
			}
			else
				S.s[i]=temp-‘0‘;
		}
		S.val=calcantor(S.s);
		puts(!bfs()?"unsolvable":ans.c_str());
	}
	return 0;
}