hdu 4801 Pocket Cube 四面魔方，模拟+搜索

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魔方只有4面，且深度只有7层，所以可以搜索解决。由于模拟过程复杂，干脆用一个辅助数组记录每次转动每个点的对应位置，在脑中空间想象一下就好了。

值得注意的一点是不需要模拟12次（6条边，顺+逆2次）因为一条边顺时针就相当于另一条边逆时针，所以只需要6次就好。

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <map>#define inf 0x3f3f3f3fusing namespace std;int ans;int n;int a[24];void debug(){    printf("      %3d%3d\n",a[0],a[1]);    printf("      %3d%3d\n",a[2],a[3]);    for(int i=4;i<=9;i++) printf("%3d",a[i]);    puts("");    for(int i=10;i<=15;i++) printf("%3d",a[i]);    puts("");    for(int j=0;j<=3;j++)    printf("      %3d%3d\n",a[16+j*2],a[16+j*2+1]);    puts("---------------");}int h[10][24]={    {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},    {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},    {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23},    {6,1,12,3,5,11,16,7,8,9,4,10,18,13,14,15,20,17,22,19,0,21,2,23},};int f(int x,int b,int c,int d){    return a[x]==a[b]&&a[b]==a[c]&&a[c]==a[d];}int cal(){    return f(0,1,2,3)+f(4,5,10,11)+f(6,7,12,13)+f(8,9,14,15)+f(16,17,18,19)+f(20,21,22,23);}void dfs(int last){    ans=max(ans,cal());    if(ans==6) return;    if(!last) return;    int tmp[24];    memcpy(tmp,a,sizeof(a));    for(int d=0;d<6;d++)    {        for(int i=0;i<24;i++) a[i]=tmp[h[d][i]];        dfs(last-1);        memcpy(a,tmp,sizeof(tmp));    }}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<24;i++)        {            scanf("%d",&a[i]);        }        ans=cal();        dfs(n);        printf("%d\n",ans);    }    return 0;}

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