ZOJ 3736 Pocket Cube

模拟。。。就足够了

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 thanN 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

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int mf[24],ans=0;bool _ck(int a,int b,int c,int d){    if(mf[a]==mf[b]&&mf[a]==mf[c]&&mf[a]==mf[d]) return true;    return false;}bool ck(){    if(_ck(0,1,2,3)&&_ck(4,5,10,11)&&_ck(6,7,12,13)&&_ck(8,9,14,15)&&_ck(16,17,18,19)&&_ck(20,21,22,23))        return true;    return false;}int count_ck(){    return _ck(0,1,2,3)+_ck(4,5,10,11)+_ck(6,7,12,13)+_ck(8,9,14,15)+_ck(16,17,18,19)+_ck(20,21,22,23);}bool zhuan(int a,int b,int c,int d,    int e,int f,   int g,int h,   int i,int j,   int k,int l){    ///圈    int t=mf[a];    mf[a]=mf[d];mf[d]=mf[c];mf[c]=mf[b];mf[b]=t;    ///环    int t1=mf[e],t2=mf[f];    mf[e]=mf[g];mf[f]=mf[h];    mf[g]=mf[i];mf[h]=mf[j];    mf[i]=mf[k];mf[j]=mf[l];    mf[k]=t1;mf[l]=t2;}void qian1(){    zhuan(13,12,6,7,3,2,5,11,16,17,14,8);}void qian2(){    zhuan(6,12,13,7,2,3,8,14,17,16,11,5);}void zhuo1(){    zhuan(15,14,8,9,1,3,7,13,17,19,21,23);}void zhuo2(){    zhuan(8,14,15,9,23,21,19,17,13,7,3,1);}void shang1(){    zhuan(16,17,19,18,20,21,15,14,13,12,11,10);}void shang2(){    zhuan(16,18,19,17,10,11,12,13,14,15,21,20);}void dfs(int k){    if(ans==6) return ;    ans=max(ans,count_ck());    if(!k) return ;    qian1();    dfs(k-1);    qian2();    qian2();    dfs(k-1);    qian1();    zhuo1();    dfs(k-1);    zhuo2();    zhuo2();    dfs(k-1);    zhuo1();    shang1();    dfs(k-1);    shang2();    shang2();    dfs(k-1);    shang1();}int main(){    int k;while(scanf("%d",&k)!=EOF){    for(int i=0;i<24;i++) scanf("%d",mf+i);    ans=count_ck();    dfs(k);    printf("%d\n",ans);}    return 0;}