Pocket Cube HDU

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<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>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;}