2012 Asia Changsha Regional Contest Problem K

Pocket Cube

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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这题写的真的*痛- -#，全局数组变了全局传过去的tmp数组也变了不知道，见识太少- -#~

因为2阶魔方本来有12中转每个面有2中方向，但是右边顺时针和左边逆时针是相同的所以其实只有6种转法！

预处理下转动后的情况就行了！

如果考虑翻转的话可以只有4种，但是不好判重，翻转又不算次数，所以还是直接6种爆搜好，标记啊，判重啊什么的，一个是为了减少次数，另一方面可以避免死循环(-｡-;)！

不过这题直接爆就行了！

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 C_i seperated by a sinle space in the second line. For index 0 ≤ i < 24, C_i 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

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Author: FAN, Yuzhe;CHEN, Cong;GUAN, Yao Contest: The 2013 ACM-ICPC Asia Changsha Regional Contest

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 100010
int n, m, s, x, d, ans;
int a[31];
int p[6][24] = {
    {6,1,12,3,5,11,16,7,8,9,4,10,18,13,14,15,20,17,22,19,0,21,2,23},
    {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23},
    {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8},
    {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},
    {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},
    {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23},
};
int judge(int *tmp){
    int t = 0;
    if (tmp[0] == tmp[1] && tmp[1] == tmp[2] && tmp[2] == tmp[3])t++;
    if (tmp[4] == tmp[5] && tmp[5] == tmp[10] && tmp[10] == tmp[11])t++;
    if (tmp[6] == tmp[7] && tmp[7] == tmp[12] && tmp[12] == tmp[13])t++;
    if (tmp[8] == tmp[9] && tmp[9] == tmp[14] && tmp[14] == tmp[15])t++;
    if (tmp[16] == tmp[17] && tmp[17] == tmp[18] && tmp[18] == tmp[19])t++;
    if (tmp[20] == tmp[21] && tmp[21] == tmp[22] && tmp[22] == tmp[23])t++;
    return t;
}
void dfs(int *tmp,int k){
    int f[41];
    ans = max(ans, judge(tmp));
    if (k==n)return;
    for (int i = 0; i < 6; i++){
        for (int j = 0; j < 24; j++)f[j] = tmp[p[i][j]]; 
        dfs(f,k + 1);
    }
}
int main(){
    while (~scanf("%d", &n)){
        ans = 0;
        for (int i=0;i<24;i++)scanf("%d", &a[i]);
        dfs(a,0);
        printf("%d\n", ans);
    }
    return 0;
}