[dfs] zoj 3736 Pocket Cube

题目链接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3736

Pocket Cube

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

10 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 510 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

62

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

题目意思：

给2*2*2的魔方，求在给定的n步内最多可以凑成多少完整的面。

解题思路：

由于n<=7,所以暴搜即可。

每种状态下有6种选择，（不是12种，因为转动右边可以转化为转动左边，转动下面可以转化为转动上面，转动后面可以转化为转动前面）

所以只用考虑左边、下面、后面的各两种情况。

找出各位置新的状态下在原来的位置标号。

Tips: 当数组指针作为参数传给函数时，不能在里面用memcpy,因为此时并不知道该数组的大小

当全局数组指针作为参数传给函数时，每个递归函数里面对指针所指内容所做的修改，都会生效，并且有可能访问的都是同一个内存区域。

当传数组指针时，最好使用局部数组这样才不会有问题。

代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 30int cur[Maxn],vv[Maxn],n,ans;int next[][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},//左边往内    {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23},//左边往外    {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},//下面往内    {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23},//下面往外    {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},//后面往下    {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8} //后面忘上};int cal(int * cur) //统计完整面的个数{    int res=0;    if(cur[0]==cur[1]&&cur[0]==cur[2]&&cur[0]==cur[3])        res++;    if(cur[4]==cur[5]&&cur[4]==cur[10]&&cur[4]==cur[11])        res++;    if(cur[6]==cur[7]&&cur[6]==cur[12]&&cur[6]==cur[13])        res++;    if(cur[8]==cur[9]&&cur[8]==cur[14]&&cur[8]==cur[15])        res++;    if(cur[16]==cur[17]&&cur[16]==cur[18]&&cur[16]==cur[19])        res++;    if(cur[20]==cur[21]&&cur[20]==cur[22]&&cur[20]==cur[23])        res++;    return res;}void dfs(int num,int * cur){    ans=max(ans,cal(cur));    if(num>n)        return ;    int vv[Maxn]; //该数组定义成全局变量会有问题    for(int i=0;i<6;i++)    {        for(int j=0;j<24;j++)            vv[j]=cur[next[i][j]]; //如果vv是全局的，vv和cur会指向同一块内存，访问会出现混乱        dfs(num+1,vv);    }}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   while(~scanf("%d",&n))   {       for(int i=0;i<24;i++)            scanf("%d",&cur[i]);       ans=0;       dfs(1,cur);       printf("%d\n",ans);   }   return 0;}