Cubes

题意

给出$12$根长度相等的木棒,颜色最多有$6$种,问能构成的本质不同的正方体数量.

题解

根据波利亚定理公式:

设X是元素集合,G是X的置换群,{u1,u2,...,uk}\{u_1,u_2,...,u_k\}{u1​,u2​,...,uk​}是$k$种颜色的集合,$C$是$X$的任意着色集.这时,针对各颜色的数目的C的非等价着色数的生成函数是:
$P_G(u_1+...+u_k,u_1^2+...+u_k^2,...,u_1^n+...+u_k^n)$

0. 我们先求出$12$根木棒的颜色数,假设有$k$中,每种颜色假设有{c1,c2,...,ck}\{c_1,c_2,...,c_k\}{c1​,c2​,...,ck​}个,注意$c_1+...+c_k=12$.

1. 求出置换群.
   众所周知的,正方体的置换群大小是$24$.
   (1)恒等变换:$1$个
   (2)绕对面的中心旋转$90,180,270$度:$3\ast 3=9$个.
   (3)绕对边的中点连成的轴旋转$180$度:$6\ast 1=6$个.
   (4)绕体对角线旋转$120,240$度:$4\ast 2=8$个.
   根据上述四种情况得到的关于线元素的置换平均值为:
   $P_G=\frac{1}{24}(z_1^{12}+6z_4^3+3z_2^6+8z_3^4+6z_1^2{z}_2^5)$

2. 将$z_i=\sum u_1^i{u}_2^i...u_6^i$代入$P_G$式.那么答案就是$P_G$函数的$u_1^{c_1}{u}_2^{c_2}...u_k^{c_k}$前的系数

3. $P_G$函数的每一项都是形如$(u_1+...+u_k{)}^a(u_1^2+...+u_k^2{)}^b...(u_1^n+...+u_k^n{)}^z$的,我要对每个这样的式子,求$u_1^a{u}_2^b...u_k^z$前的系数之和.每个项单独进行$dfs$就可以了.

代码

#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)int T;
int tc[7],x[7],ox[7];
int C[13][13];void init() {C[0][0] = 1;rep(i,1,12) {C[i][0] = 1;rep(j,1,i) {C[i][j] = C[i-1][j] + C[i-1][j-1];}}
}
int ans = 0;
void dfs(int dep,int mul) {if(dep == 5) {int f = 0;rep(i,1,6) f += x[i];if(!f) {ans += mul;}return ;}int tx[7];rep(i,1,6) tx[i] = x[i];for(int x1 = 0;x1 <= tc[dep] && dep * x1 <= tx[1];++x1) {int res1 = C[tc[dep]][x1];for(int x2 = 0;x2 <= tc[dep] - x1 && dep*x2 <= tx[2];++x2) {int res2 = res1 * C[tc[dep]-x1][x2];for(int x3 = 0;x3 <= tc[dep]-x1-x2 && dep*x3 <= tx[3];++x3) {int res3 = res2 * C[tc[dep]-x1-x2][x3];for(int x4 = 0;x4 <= tc[dep]-x1-x2-x3 && dep*x4 <= tx[4];++x4) {int res4 = res3 * C[tc[dep]-x1-x2-x3][x4];for(int x5 = 0;x5 <= tc[dep]-x1-x2-x3-x4 && dep*x5 <= tx[5];++x5) {int res5 = res4 * C[tc[dep]-x1-x2-x3-x4][x5];int x6 = tc[dep]-x1-x2-x3-x4-x5;//pr(x1);pr(x2);pr(x3);pr(x4);pr(x5);pr(x6);if(dep*x6 <= tx[6]) {x[1] = tx[1] - dep*x1;x[2] = tx[2] - dep*x2;x[3] = tx[3] - dep*x3;x[4] = tx[4] - dep*x4;x[5] = tx[5] - dep*x5;x[6] = tx[6] - dep*x6;dfs(dep+1,res5*mul);}}}}}}
}int main() {init();std::ios::sync_with_stdio(false);std::cin >> T;while(T--) {memset(ox,0,sizeof(ox));rep(i,1,12) {int tmp;std::cin >> tmp;ox[tmp] ++;}int res = 0;rep(i,1,6) x[i] = ox[i];ans = 0;tc[1] = 12,tc[2] = tc[3] = tc[4] = 0;dfs(1,1);res += ans;rep(i,1,6) x[i] = ox[i];ans = 0;tc[1] = tc[2] = tc[3] = 0,tc[4] = 3;dfs(1,1);res += 6*ans;rep(i,1,6) x[i] = ox[i];ans = 0;tc[1] = tc[3] = tc[4] = 0,tc[2] = 6;dfs(1,1);res += 3*ans;rep(i,1,6) x[i] = ox[i];ans = 0;tc[1] = tc[2] = tc[4] = 0,tc[3] = 4;dfs(1,1);res += 8*ans;rep(i,1,6) x[i] = ox[i];ans = 0;tc[1] = 2,tc[2] = 5,tc[3] = tc[4] = 0;dfs(1,1);res += 6*ans;std::cout << res/24 << std::endl;}return 0;
}