hdu 5012 Dice

Problem Description

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.

 

Input

There are multiple test cases. Please process till EOF. 
For each case, the first line consists of six integers a₁,a₂,a₃,a₄,a₅,a₆, representing the numbers on dice A. 
The second line consists of six integers b₁,b₂,b₃,b₄,b₅,b₆, 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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

题意：两个骰子，要把第一个骰子转到和第二个一样，有4种转法，求最少的次数

直接bfs，不过我用G++提交超时，用C++提交可以过，可能是使用了queue的原因。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
struct Node
{
    int a[7];
    int t;
}st,ed;
int vis[10][10][10][10][10][10];
void bfs()
{
    
    queue<Node>q;
    q.push(st);
    
    vis[st.a[0]][st.a[1]][st.a[2]][st.a[3]][st.a[4]][st.a[5]]=1;
    Node t1,t2;
    Node tmp;
    while(!q.empty())
    {
        t1=q.front();
        q.pop();
        if(t1.a[0]==ed.a[0] && t1.a[1]==ed.a[1]&& t1.a[2]==ed.a[2]&& t1.a[3]==ed.a[3]&& t1.a[4]==ed.a[4] && t1.a[5]==ed.a[5])
        {
            printf("%d
",t1.t);
            return;
        }
        
        t2=t1;
        t2.a[0]=t1.a[3];
        t2.a[1]=t1.a[2];
        t2.a[2]=t1.a[0];
        t2.a[3]=t1.a[1];
        if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0)
        {
            vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1;
            t2.t=t1.t+1;
            q.push(t2);
        }
        
        
        t2=t1;
        t2.a[0]=t1.a[2];
        t2.a[1]=t1.a[3];
        t2.a[2]=t1.a[1];
        t2.a[3]=t1.a[0];
        if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0)
        {
            vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1;
            t2.t=t1.t+1;
            q.push(t2);
        }
        
        t2=t1;
        t2.a[0]=t1.a[5];
        t2.a[1]=t1.a[4];
        t2.a[4]=t1.a[0];
        t2.a[5]=t1.a[1];
        if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0)
        {
            vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1;
            t2.t=t1.t+1;
            q.push(t2);
        }
        
        t2=t1;
        t2.a[0]=t1.a[4];
        t2.a[1]=t1.a[5];
        t2.a[4]=t1.a[1];
        t2.a[5]=t1.a[0];
        if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0)
        {
            vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1;
            t2.t=t1.t+1;
            q.push(t2);
        }
    }
    printf("-1
");
}
int main()
{
    while(scanf("%d%d%d%d%d%d",&st.a[0],&st.a[1],&st.a[2],&st.a[3],&st.a[4],&st.a[5])==6)
    {
        scanf("%d%d%d%d%d%d",&ed.a[0],&ed.a[1],&ed.a[2],&ed.a[3],&ed.a[4],&ed.a[5]);
        st.t=0;
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}