UVA - 10601 Cubes （组合+置换）

Description

Problem B

Cubes

You are given 12 rods of equal length. Each of them is colored in certain color. Your task is to determine in how many different ways one can construct a cube using these rods as edges. Two cubes are considered equal if one of them could be rotated and put
next to the other, so that the corresponding edges of the two cubes are equally colored.

Input

The first line of input contains T (1≤T≤60), the number of test cases. Then T test cases follow. Each test case consists of one line containing 12 integers. Each of them denotes the color of the corresponding rod. The colors are numbers between 1 and 6.

Output

The output for one test consists of one integer on a line - the number of ways one can construct a cube with the described properties.

Sample Input	Sample Output
3 1 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 3 3 4 4 5 5 6 6	1 5 312120

Problem source: Bulgarian National Olympiad in Informatics 2003

Problem submitter: Ivaylo Riskov

Problem solution: Ivaylo Riskov, K M Hasan

题意：有12条边，每个边有指定的颜色，组成一个立方体，求种数

思路：参看cxlove GG的博客

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int maxn = 20;

int a[maxn], b[maxn];
ll c[maxn][maxn];

void init() {
	for (int i = 0; i <= 12; i++) {
		c[i][0] = c[i][i] = 1;
		for (int j = 1; j < i; j++)
			c[i][j] += c[i-1][j] + c[i-1][j-1];
	}
}

ll solve(int k) {
	ll sum = 1;
	int n = 0;
	for (int i = 0; i < 6; i++)
		if (b[i] % k == 0) {
			b[i] /= k;
			n += b[i]; 
		}
		else return 0;

	for (int i = 0; i < 6; i++) {
		sum *= c
[b[i]];
		n -= b[i];
	}
	return sum;
}

ll still() {
	memcpy(b, a, sizeof(a));
	return solve(1);
}

ll point() {
	memcpy(b, a, sizeof(a));
	return 4*2*solve(3);
}

ll plane() {
	memcpy(b, a, sizeof(a));
	ll ans = 3*2*solve(4);
	memcpy(b, a, sizeof(a));
	return ans + 3*solve(2);
}

ll edge() {
	ll ans = 0;
	for (int i = 0; i < 6; i++)
		for (int j = 0; j < 6; j++) {
			memcpy(b, a, sizeof(b));
			b[i]--, b[j]--;
			if (b[i] < 0 || b[j] < 0)
				continue;
			ans += 6 * solve(2);
		}
	return ans;
}

ll polya() {
	ll ans = 0;
	ans += still();
	ans += point();
	ans += plane();
	ans += edge();
	return ans / 24;
}

int main() {
	init();
	int t, k;
	scanf("%d", &t);
	while (t--) {
		memset(a, 0, sizeof(a));
		for (int i = 0; i < 12; i++) {
			scanf("%d", &k);
			a[k-1]++;
		}
		printf("%lld\n", polya());
	}
	return 0;
}