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 0x3f3f3f3f using 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; }