2013区域赛长沙赛区现场赛 K - Pocket Cube

K - Pocket Cube

Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4801

Description

Pocket Cube is a 3-D combination puzzle. It is a 2 × 2 × 2 cube, which means it is constructed by 8 mini-cubes. For a combination of 2 × 2 mini-cubes which sharing a whole cube face, you can twist it 90 degrees in clockwise or counterclockwise direction, this twist operation is called one twist step. 
Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes' faces of same color rely on same large cube face, we can call the large cube face as a completed face.

Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps. 
Index of each face is shown as below:

 

Input

There will be several test cases. In each test case, there will be 2 lines. One integer N (1 ≤ N ≤ 7) in the first line, then 24 integers Ci separated by a single space in the second line. For index 0 ≤ i < 24, Ci is color of the corresponding face. We guarantee that the color arrangement is a valid state which can be achieved by doing a finite number of twist steps from an initial cube whose all 6 large cube faces are completed faces.

 

Output

For each test case, please output the maximum number of completed faces during no more than N twist step(s).

 

Sample Input

1
0 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 5
1
0 4 0 4 1 1 2 5 3 3 1 1 2 5 3 3 4 0 4 0 5 2 5 2

 

Sample Output

6
2

 

转魔方，bfs，一直找不到wa在哪里。这题思考魔方的旋转有点烦。

有6种旋转方式，而不是12种。。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<vector>
 #include<queue>
 #include<algorithm>
 #include<map>
 #define INF 0x3f3f3f3f
 #define M(a,b) memset(a,b,sizeof(a))
 
 using namespace std;
 
 int state[];
 int change[][]={
     {,,,,,,,,,,,,,,,,,,,,,,,},
     {,,,,,,,,,,,,,,,,,,,,,,,},
     {,,,,,,,,,,,,,,,,,,,,,,,},
     {,,,,,,,,,,,,,,,,,,,,,,,},
     {,,,,,,,,,,,,,,,,,,,,,,,},
     {,,,,,,,,,,,,,,,,,,,,,,,},
 };
 
 int N;
 int ans;
 
 struct tube
 {
     int d[];
     int cnt;
     tube(){}
 };
 tube st;
 tube trans(tube tmp,int k)
 {
     tube res;
 
     for (int i=; i<; i++)
     res.d[i]=tmp.d[change[k][i]];
 
     return res;
 }
 int count(tube tmp)
 {
     int res=;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     if (tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[] && tmp.d[]==tmp.d[]) res++;
     return res;
 }
 
 void bfs()
 {
     queue<tube> q;
     q.push(st);
 //    map<tube,int> mp;
 //    mp[st]=1;
     ans=count(st);
     while(!q.empty())
     {
         tube now=q.front();
         q.pop();
         tube tmp;
         for (int i=; i<; i++)
         {
             tmp=trans(now,i);
             tmp.cnt=now.cnt+;
 //            if (mp.find(tmp)!=mp.end())
 //            {
 //                mp[tmp]=1;
 //                ans=max(ans,count(tmp));
 //                if (tmp.cnt<num) q.push(tmp);
 //            }
             ans=max(ans,count(tmp));
             if (tmp.cnt<N) q.push(tmp);
         }
     }
 }
 
 int main()
 {
    //init();
    while(scanf("%d",&N)==)
    {
        for(int i = ;i<;i++)
        {
             scanf("%d",&st.d[i]);
         }
         st.cnt=;
         bfs();
        printf("%d\n",ans);
    }
    return ;
 }