【BZOJ 1004】【HNOI 2008】Cards

http://www.lydsy.com/JudgeOnline/problem.php?id=1004

注意数据给出的m是一个没有单位元的置换群！

用Burnside引理，然后对每个置换群dp一下就可以了。

#include<cstdio>
#include<bitset>
#include<cstring>
#include<algorithm>
using namespace std;

int a[63], Sr, Sb, Sg, m, p, n, ans = 0;

int ipow(int w, int b) {
	int ret = 1;
	while (b) {
		if (b & 1) ret = ret * w % p;
		w = w * w % p;
		b >>= 1;
	}
	return ret;
}

int f[63][63][63][63], w[63], tot;
bitset <64> vis;

int dp() {
	vis.reset(); tot = 0;
	for (int i = 1; i <= n; ++i)
		if (!vis[i]) {
			vis[i] = 1;
			w[++tot] = 1;
			int tmp = a[i];
			while (!vis[tmp]) {
				vis[tmp] = 1;
				++w[tot];
				tmp = a[tmp];
			}
		}

	for (int i = 1; i <= tot; ++i)
		for (int R = 0; R <= Sr; ++R)
			for (int B = 0; B <= Sb; ++B)
				for (int G = 0; G <= Sg; ++G) {
					f[i][R][B][G] = 0;
					if (R >= w[i]) (f[i][R][B][G] += f[i - 1][R - w[i]][B][G]) %= p;
					if (B >= w[i]) (f[i][R][B][G] += f[i - 1][R][B - w[i]][G]) %= p;
					if (G >= w[i]) (f[i][R][B][G] += f[i - 1][R][B][G - w[i]]) %= p;
				}
	return f[tot][Sr][Sb][Sg];
}

int main() {
	scanf("%d%d%d%d%d", &Sr, &Sb, &Sg, &m, &p);
	n = Sr + Sb + Sg; f[0][0][0][0] = 1;
	for (int j = 1; j <= n; ++j) a[j] = j;
	(ans += dp()) %= p;
	for (int i = 1; i <= m; ++i) {
		for (int j = 1; j <= n; ++j) scanf("%d", a + j);
		(ans += dp()) %= p;
	}
	printf("%d\n", ans * ipow(m + 1, p - 2) % p);
	return 0;
}