Pocket Cube HDU - 4801 (dfs)

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
直接暴力模拟，最近发现dfs的题都写不出

#include
#include
#include
#include
#include
using namespace std;
int tran[7][24]={
                {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23},
                {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},
                {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},
                {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},
                };
int M[24];
int ans=0;
int n;
void dfs(int step,int *a)
{
    int num=0;
    if(a[4]==a[5]&&a[4]==a[10]&&a[4]==a[11])
        num++;
    if(a[6]==a[13]&&a[6]==a[7]&&a[6]==a[12])
        num++;
    if(a[8]==a[9]&&a[8]==a[14]&&a[8]==a[15])
        num++;
    if(a[0]==a[1]&&a[0]==a[2]&&a[0]==a[3])
        num++;
    if(a[16]==a[17]&&a[16]==a[18]&&a[16]==a[19])
        num++;
    if(a[20]==a[21]&&a[20]==a[23]&&a[20]==a[22])
        num++;
        ans=max(ans,num);
    if(step==n||ans==6)
        return ;
    int m[24];
    for(int i=0;i<6;i++)
    {
        for(int j=0;j<24;j++)
            m[j]=a[tran[i][j]];
        dfs(step+1,m);
    }
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<24;i++)
            scanf("%d",M+i);
            ans=0;
        dfs(0,M);
        printf("%d\n",ans);

    }
    return 0;
}