HDU - 5012 Dice

Dice

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a 1.a 2,a 3,a 4,a 5,a 6 to
be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b 1.b 2,b 3,b 4,b 5,b 6 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.



Input

There are multiple test cases. Please process till EOF.

For each case, the first line consists of six integers a 1,a 2,a 3,a 4,a 5,a 6, representing the numbers on dice
A.

The second line consists of six integers b 1,b 2,b 3,b 4,b 5,b 6, representing the numbers on dice B.


Output

For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.


Sample Input

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

Sample Output

0
3
-1

#include<iostream>
#include<queue>
#include<string>
#include<cstdlib>
#include<cstdio>
using namespace std;         //bfs,很水很水的题
int a[7];
int b[7];
int v[7][7][7][7][7][7];
struct node
{
	int d[7];
	int ans;
};
struct node s, t;
int bfs()
{
	queue<node>q;
	for (int i = 1; i <= 6; i++)
		t.d[i] = a[i];
	t.ans = 0;
	q.push(t);
	v[t.d[1]][t.d[2]][t.d[3]][t.d[4]][t.d[5]][t.d[6]] = 1;
	while (!q.empty())
	{
		t = q.front();
		q.pop();
		if (t.d[1] == b[1] && t.d[2] == b[2] && t.d[3] == b[3] && t.d[4] == b[4] && t.d[5] == b[5] && t.d[6] == b[6])
		{
			printf("%d\n", t.ans);
			return 1;
		}
		for (int i = 1; i <= 4; i++)
		{
			if (i == 1)
			{
				s.d[1] = t.d[4];
				s.d[2] = t.d[3];
				s.d[3] = t.d[1];
				s.d[4] = t.d[2];
				s.d[5] = t.d[5];
				s.d[6] = t.d[6];
				if (v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] == 0)
				{
					s.ans = t.ans + 1;
					v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] = 1;
					q.push(s);
				}
			}
			else if (i == 2)
			{
				s.d[1] = t.d[3];
				s.d[2] = t.d[4];
				s.d[3] = t.d[2];
				s.d[4] = t.d[1];
				s.d[5] = t.d[5];
				s.d[6] = t.d[6];
				if (v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] == 0)
				{
					s.ans = t.ans + 1;
					v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] = 1;
					q.push(s);
				}
			}
			else if (i == 3)
			{
				s.d[1] = t.d[6];
				s.d[2] = t.d[5];
				s.d[3] = t.d[3];
				s.d[4] = t.d[4];
				s.d[5] = t.d[1];
				s.d[6] = t.d[2];
				if (v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] == 0)
				{
					s.ans = t.ans + 1;
					v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] = 1;
					q.push(s);
				}
			}
			else if (i == 4)
			{
				s.d[1] = t.d[5];
				s.d[2] = t.d[6];
				s.d[3] = t.d[3];
				s.d[4] = t.d[4];
				s.d[5] = t.d[2];
				s.d[6] = t.d[1];
				if (v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] == 0)
				{
					s.ans = t.ans + 1;
					v[s.d[1]][s.d[2]][s.d[3]][s.d[4]][s.d[5]][s.d[6]] = 1;
					q.push(s);
				}
			}
		}
	}
	return 0;
}

int main()
{
	int F;
	while (scanf("%d%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6])!=EOF) //加EOF啊笨蛋，在这TE这么多次太对不起学的算法了
	{   
		for (int i = 1; i <= 6; i++)
			scanf("%d", &b[i]);
		memset(v, 0, sizeof(v));
		F = bfs();
		if (F==0) printf("-1\n");
	}
}