[洛谷P4208][JSOI2008]最小生成树计数

题目大意：有$n$个点和$m$条边（最多有$10$条边边权相同），求最小生成树个数

题解：对于所有最小生成树，每种边权的边数是一样的。于是就可以求出每种边权在最小生成树中的个数，枚举这种边的边集，求出对于这个边集可以的解（即没有一条边在同一联通块中），再把每种边的方案数乘起来即可。

卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
using namespace std;
const long long mod = 31011;
struct Edge {
	int from, to, w;
	bool operator < (const Edge &a) const {return w < a.w;}
} e[1010];
int n, m;
struct Set {
	int f[111];
	int find(int x) {return ((x == f[x]) ? x : (f[x] = find(f[x])));}
	bool operator = (const Set &a) {
		for (int i = 1; i <= n; i++) f[i] = a.f[i];
	}
} s1, s2;
long long ans = 1;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) scanf("%d%d%d", &e[i].from, &e[i].to, &e[i].w);
	sort(e + 1, e + m + 1);
	for (int i = 1; i <= n; i++) s1.f[i] = s2.f[i] = i;
	for (int i = 1; i <= m; i++) {
		int same = i, cnt = 0, res = 0;
		while (same < m && e[same].w == e[same + 1].w) same++;
		s1 = s2;
		for (int j = i; j <= same; j++) {
			int u = s1.find(e[j].from), v = s1.find(e[j].to);
			if (u != v) {
				s1.f[u] = v;
				cnt++;
			}
		}
		for (int j = 0; j < 1 << same - i + 1; j++) {
			bool flag = false;
			if (__builtin_popcount(j) == cnt) {
				flag = true;
				s1 = s2;
				for (int k = i; k <= same; k++) {
					if (j & 1 << k - i) {
						int u = s1.find(e[k].from), v = s1.find(e[k].to);
						if (u == v) {
							flag = false;
							break;
						} else s1.f[u] = v;
					}
				}
			}
			res += flag;
		}
		for (int j = i; j <= same; j++) {
			int u = s2.find(e[j].from), v = s2.find(e[j].to);
			if (u != v) s2.f[u] = v;
		}
		ans = (ans * res) % mod;
	}
	int tmp = s2.find(1);
	for (int i = 2; i <= n; i++) if (s2.find(i) != tmp) {
		puts("0");
		return 0;
	}
	printf("%d
", ans);
	return 0;
}