K - Pocket Cube

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

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 ;
}

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