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

题目及代码：

       直接光搜就可以了，最多搜索到5次就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

struct node
{
    int s[6];
    int step;
}l,r,t,tmp;

queue<node>Q;

bool judge(node t)
{
    for(int i=0;i<6;i++)
    if(t.s[i]!=r.s[i]) return false;
    return true;
}

void left(node &t)
{
    int tmp=t.s[3];
    t.s[3]=t.s[1];
    t.s[1]=t.s[2];
    t.s[2]=t.s[0];
    t.s[0]=tmp;
}

void right(node &t)
{
    int tmp=t.s[0];
    t.s[0]=t.s[2];
    t.s[2]=t.s[1];
    t.s[1]=t.s[3];
    t.s[3]=tmp;
}

void front(node &t)
{
    int tmp=t.s[5];
    t.s[5]=t.s[1];
    t.s[1]=t.s[4];
    t.s[4]=t.s[0];
    t.s[0]=tmp;
}

void back(node &t)
{
    int tmp=t.s[0];
    t.s[0]=t.s[4];
    t.s[4]=t.s[1];
    t.s[1]=t.s[5];
    t.s[5]=tmp;
}

int bfs()
{
    while(!Q.empty())
    {
        Q.pop();
    }
    Q.push(l);
    while(!Q.empty())
    {
        t=Q.front();
        Q.pop();
        if(t.step>=5) break;
        if(judge(t)) return t.step;


        t.step++;
        tmp=t;
        right(tmp);
        Q.push(tmp);

        tmp=t;
        left(tmp);
        Q.push(tmp);

        tmp=t;
        front(tmp);
        Q.push(tmp);

        tmp=t;
        back(tmp);
        Q.push(tmp);
    }
    return -1;
}
int main()
{
    while(scanf("%d",&l.s[0])!=EOF)
    {
        l.step=0;
        for(int i=1;i<6;i++)
        {
            scanf("%d",&l.s[i]);
        }
        for(int i=0;i<6;i++)
        {
            scanf("%d",&r.s[i]);
        }
        int n=bfs();
        printf("%d\n",n);
    }

    return 0;
}

hdu 5012 Dice

原文：http://blog.csdn.net/knight_kaka/article/details/39298517