使用带扩展域的并查集，将每个节点 $p$ 拆分为 $p_{friend}$ 和 $p_{enemy}$ 两个域，最后统计 $1\sim n$ 的祖先个数即可。

#include <iostream>
#include <bitset>

using namespace std;

bitset<4005> v;
int fa[2005];
int n, m, ans;

int find(int x) {
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void unite(int x, int y) {
	fa[find(x)] = find(y);
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);

	cin >> n >> m;
	for (int i = 1; i <= n * 2; ++i) fa[i] = i;
	while (m--) {
		char opt;
		int p, q;
		cin >> opt >> p >> q;
		int pf = p, pe = p + n, qf = q, qe = q + n;
		if (opt == 'F') unite(pf, qf);
		else unite(pe, qf), unite(pf, qe);
	}
	for (int i = 1; i <= n; ++i) {
		int root = find(i);
		if (!v[root]) v[root] = true, ++ans;
	}
	cout << ans << endl;

	return 0;
}