[arc062e]Building Cubes with AtCoDeer
题意：
给出N个正方形，每个正方形正面的四个角各有一种颜色，要求组成不同的正方体的方案数。两种方案视为相同，当且仅当一个正方体可以通过各种旋转得到另一个。（方向不同也算不同方案）
$6\leq N\leq 400$
题解：
输出样例+puts("0")就可以得到40分的好成绩！
然后这是一道暴力题！
那么写个暴力就ok了！
1 #include<algorithm>
2 #include<iostream>
3 #include<cstring>
4 #include<cstdio>
5 #include<cmath>
6 #include<unordered_map>
7 #include<map>
8 using namespace std;
9 typedef long long ________________________;
10 typedef int _________________________;
11 const ________________________ _=1000,__=1000000,___=1000000000;
12 _________________________ ____,_____[401][4];
13 unordered_map<________________________,________________________>______;
14 ________________________ _______=0,________,_________[4];
15 ________________________ __________(_________________________ __________________,_________________________ ___________________,_________________________ ____________________,_________________________ _____________________){
16 return (________________________)(__________________+___________________*_+____________________*__+_____________________*___);
17 }
18 void ____________(_________________________ __________________,_________________________ ___________________,_________________________ ____________________,_________________________ _____________________,_________________________ _______________________){
19 ______[__________(__________________,___________________,____________________,_____________________)]+=_______________________;
20 ______[__________(___________________,____________________,_____________________,__________________)]+=_______________________;
21 ______[__________(____________________,_____________________,__________________,___________________)]+=_______________________;
22 ______[__________(_____________________,__________________,___________________,____________________)]+=_______________________;
23 }
24 void _____________(________________________ ______________________,_________________________ _______________________){
25 _________________________ __________________,___________________,____________________,_____________________;
26 __________________=______________________%_,______________________/=_,___________________=______________________%_,______________________/=_,____________________=______________________%_,______________________/=_,_____________________=______________________%_,______________________/=_;
27 ____________(__________________,___________________,____________________,_____________________,_______________________);
28 }
29 _________________________ main(){
30 scanf("%d",&____);
31 for(_________________________ ______________=1;______________<=____;______________++){
32 scanf("%d%d%d%d",&_____[______________][0],&_____[______________][1],&_____[______________][2],&_____[______________][3]);
33 ____________(_____[______________][0],_____[______________][1],_____[______________][2],_____[______________][3],1);
34 }
35 for(_________________________ ______________=1;______________<____;______________++){
36 ____________(_____[______________][0],_____[______________][1],_____[______________][2],_____[______________][3],-1);
37 for(_________________________ _______________=______________+1;_______________<=____;_______________++){
38 ____________(_____[_______________][0],_____[_______________][1],_____[_______________][2],_____[_______________][3],-1);
39 for(_________________________ ________________=0;________________<4;________________++){
40 ________=1;
41 for(_________________________ _________________=0;_________________<4;_________________++){
42 _________[_________________]=__________(_____[______________][_________________],_____[_______________][(________________+5-_________________)&3],_____[_______________][(________________+4-_________________)&3],_____[______________][(_________________+1)&3]);
43 }
44 if(!______[_________[0]]||!______[_________[1]]||!______[_________[2]]||!______[_________[3]])continue;
45 for(_________________________ _________________=0;_________________<4;_________________++){
46 ________*=______[_________[_________________]];
47 _____________(_________[_________________],-1);
48 }
49 for(_________________________ _________________=0;_________________<4;_________________++)_____________(_________[_________________],1);
50 _______+=________;
51 }
52 ____________(_____[_______________][0],_____[_______________][1],_____[_______________][2],_____[_______________][3],1);
53 }
54 }
55 printf("%lld",_______);
56 return 0;
57 }
本文到此结束！
以上是我的口胡。嗯，被这题的题解感动到语无伦次的后果。
但是题解实际上就是暴力。。。。。。注意到一个性质就是如果两个对面的正方形确定了，那么整个正方体就唯一确定了。由于n特别小，所以可以直接暴力枚举两个正方形，用map存储方案即可。注意正方形可以旋转，所以四个方向都要加入map中。
看起来很简单，但是场上只有myh过了。。。其他人大多是40$\rightarrow$_$\rightarrow$（打表出奇迹）
代码放过就不给啦！
