Dice

时间：2018-11-11 20:10:12 阅读：259 评论：0 收藏：0 [点我收藏+]

标签：clu ota and asi images ati tar case reg

Dice

http://acm.hdu.edu.cn/showproblem.php?pid=5012

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2352 Accepted Submission(s): 1178

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)

技术分享图片

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 5 6 4 3

1 2 3 4 5 6

1 4 2 5 3 6

Sample Output

-1

Source

2014 ACM/ICPC Asia Regional Xi‘an Online

模拟搜索水题

  1 #include<iostream>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<string>
  6 #include<cstdio>
  7 #include<queue>
  8 #include<map>
  9 #include<vector>
 10 typedef long long ll;
 11 #define maxn 100005
 12 using namespace std;
 13 
 14 int a[15],b[15];
 15 map<int,int>mp;
 16 struct sair{
 17     int num,step;
 18 };
 19 
 20 sair Roll(sair tmp,int x){
 21     //左右前后
 22     sair ans;
 23     ans.num=0;
 24     int aa[8];
 25     int xx=tmp.num;
 26     for(int i=6;i>=1;i--){
 27         a[i]=xx%10;
 28         xx/=10;
 29     }
 30     if(x==0){
 31         ans.num=ans.num*10+a[4];
 32         ans.num=ans.num*10+a[3];
 33         ans.num=ans.num*10+a[1];
 34         ans.num=ans.num*10+a[2];
 35         ans.num=ans.num*10+a[5];
 36         ans.num=ans.num*10+a[6];
 37     }
 38     else if(x==1){
 39         ans.num=ans.num*10+a[3];
 40         ans.num=ans.num*10+a[4];
 41         ans.num=ans.num*10+a[2];
 42         ans.num=ans.num*10+a[1];
 43         ans.num=ans.num*10+a[5];
 44         ans.num=ans.num*10+a[6];
 45     }
 46     else if(x==2){
 47         ans.num=ans.num*10+a[6];
 48         ans.num=ans.num*10+a[5];
 49         ans.num=ans.num*10+a[3];
 50         ans.num=ans.num*10+a[4];
 51         ans.num=ans.num*10+a[1];
 52         ans.num=ans.num*10+a[2];
 53     }
 54     else{
 55         ans.num=ans.num*10+a[5];
 56         ans.num=ans.num*10+a[6];
 57         ans.num=ans.num*10+a[3];
 58         ans.num=ans.num*10+a[4];
 59         ans.num=ans.num*10+a[2];
 60         ans.num=ans.num*10+a[1];
 61     }
 62     ans.step=tmp.step+1;
 63     return ans;
 64 }
 65 
 66 void bfs(int S,int E){
 67     sair s,e;
 68     if(S==E){
 69         cout<<0<<endl;
 70         return;
 71     }
 72     s.num=S,s.step=0;
 73     queue<sair>Q;
 74     Q.push(s);
 75     while(!Q.empty()){
 76         s=Q.front();
 77         Q.pop();
 78         for(int i=0;i<4;i++){
 79             e=Roll(s,i);
 80             if(e.num==E){
 81                 cout<<e.step<<endl;
 82                 return;
 83             }
 84             if(!mp[e.num]){
 85                 Q.push(e);
 86                 mp[e.num]=1;
 87             }
 88         }
 89     }
 90     cout<<-1<<endl;
 91 }
 92 
 93 int main(){
 94     while(cin>>a[1]){
 95         mp.clear();
 96         for(int i=2;i<=6;i++) cin>>a[i];
 97         int S=0,E=0;
 98         for(int i=1;i<=6;i++) E=E*10+a[i];
 99         for(int i=1;i<=6;i++) cin>>b[i];
100         for(int i=1;i<=6;i++) S=S*10+b[i];
101         bfs(S,E);
102     }
103 }

View Code

Dice

标签：clu ota and asi images ati tar case reg

原文地址：https://www.cnblogs.com/Fighting-sh/p/9943043.html

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