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 thanNtwist 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 integerN(1 ≤N≤ 7) in the first line, then 24 integersC_iseperated by a sinle space in the second line. For index 0 ≤i< 24,C_iis 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 thanNtwist 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 <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;
}