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BFS爆搜

Dice

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a₁.a₂,a₃,a₄,a₅,a₆ to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b₁.b₂,b₃,b₄,b₅,b₆ to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while a_i ≠ a_j and b_i ≠ b_j for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, a_i ≠ b_i). Ddy wants to make the two dices look the same from all directions(which means for all i, a_i = b_i) only by the following four rotation operations.(Please read the picture for more information)


Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6

0
3
-1

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

using namespace std;

struct DICE
{
	int face[6];
}A,B;

bool check(DICE a,DICE b)
{
	for(int i=0;i<6;i++)
		if(a.face[i]!=b.face[i]) return false;
	return true;
}

bool checkIMP(DICE x,DICE y)
{
	int a[6],b[6];
	for(int i=0;i<6;i++)
	{
		a[i]=x.face[i];
		b[i]=y.face[i];
	}
	sort(a,a+6);
	sort(b,b+6);
	for(int i=0;i<6;i++)
	{
		if(a[i]!=b[i]) return false;
	}
	return true;
}

void init(DICE& x)
{
	int a[6];
	for(int i=0;i<6;i++)
		a[i]=x.face[i];
	sort(a,a+6);
	for(int i=0;i<6;i++)
	{
		x.face[i]=lower_bound(a,a+6,x.face[i])-a+1;
	}
}

void showit(DICE x)
{
	for(int i=0;i<6;i++)
		cout<<x.face[i]<<",";
	cout<<endl;
}

int hash(DICE x)
{
	int ret=0;
	for(int i=0;i<6;i++)
		ret=ret*10+x.face[i];
	return ret;
}

void rhash(int h,DICE& x)
{
	for(int i=5;i>=0;i--)
	{
		x.face[i]=h%10;
		h/=10;
	}
}

void LEFT(DICE& X)
{
	int a[6];
	for(int i=0;i<6;i++)
		a[i]=X.face[i];
	X.face[0]=a[3];
	X.face[1]=a[2];
	X.face[2]=a[0];
	X.face[3]=a[1];
}

void RIGHT(DICE& X)
{
	int a[6];
	for(int i=0;i<6;i++)
		a[i]=X.face[i];
	X.face[0]=a[2];
	X.face[1]=a[3];
	X.face[3]=a[0];
	X.face[2]=a[1];
}

void UP(DICE& X)
{
	int a[6];
	for(int i=0;i<6;i++)
		a[i]=X.face[i];
	X.face[0]=a[5];
	X.face[1]=a[4];
	X.face[4]=a[0];
	X.face[5]=a[1];
}

void DOWN(DICE& X)
{	
	int a[6];
	for(int i=0;i<6;i++)
		a[i]=X.face[i];
	X.face[0]=a[4];
	X.face[1]=a[5];
	X.face[4]=a[1];
	X.face[5]=a[0];
}

bool st[1000001];

void bfs()
{
	queue<int> q,qt; 
	memset(st,0,sizeof(st));
	q.push(hash(A)); 
	st[hash(A)]=0;
	qt.push(0);
	bool flag=false;
	while(!q.empty())
	{
		int time=qt.front(); qt.pop();
		int zhi=q.front(); q.pop();
		
		DICE X,Y;int ha;
		rhash(zhi,X); Y=X;
		if(check(Y,B)==true)
		{
			flag=true;
			printf("%d\n",time);
			return ;
		}
		///LEFT
		LEFT(X);
		ha=hash(X);
		if(st[ha]==false)
		{
			st[ha]=true;
			q.push(ha);
			qt.push(time+1);
		}

		///Right
		X=Y;
		RIGHT(X);
		ha=hash(X);
		if(st[ha]==false)
		{
			st[ha]=true;
			q.push(ha);
			qt.push(time+1);
		}

		///UP
		X=Y;
		UP(X);
		ha=hash(X);
		if(st[ha]==false)
		{
			st[ha]=true;
			q.push(ha);
			qt.push(time+1);
		}

		///DOWN
		X=Y;
		DOWN(X);
		ha=hash(X);
		if(st[ha]==false)
		{
			st[ha]=true;
			q.push(ha);
			qt.push(time+1);
		}
	}
	if(flag==false) puts("-1");
}
	

int main()
{
	int x;
	while(scanf("%d",&x)!=EOF)
	{
		A.face[0]=x;
		for(int i=1;i<6;i++) scanf("%d",&A.face[i]);
		for(int i=0;i<6;i++) scanf("%d",&B.face[i]);
		
		if(checkIMP(A,B)==false)	
		{
			puts("-1"); continue;
		}
		init(A); init(B);
		if(check(A,B)==true)
		{
			puts("0"); continue;
		}
		bfs();
	}
	return 0;
}

HDOJ 5012 Dice

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原文地址：http://blog.csdn.net/ck_boss/article/details/39276619