根据每个数收那个位置初值影响建边，能证明只存在自环，但在洛谷上的 #7 出现了环

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define re register
#define sz(x) (int(x.size()))
#define all(x) x.begin(), x.end()

using namespace std;

int rd() {
	int w = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {w = (w << 3) + (w << 1) + ch - 48; ch = getchar();}
	return f * w;
}

int main() {
	int c, t;
	scanf("%d%d", &c, &t);
	while (t -- ) {
		int n = rd(), m = rd(), ans = 0;
		vector<int> fu(n + 10), p(n + 10), vis(n + 10);
		for (int i = 1; i <= n; i ++ ) fu[i] = 0, p[i] = i;
		for (int i = 1; i <= m; i ++ ) {
			char c;
			cin >> c;
			if (c == 'T') {
				int x = rd();
				fu[x] = 0;
				p[x] = n + 1;
			} else if (c == 'F') {
				int x = rd();
				fu[x] = 0;
				p[x] = n + 2;
			} else if (c == 'U') {
				int x = rd();
				fu[x] = 0;
				p[x] = n + 3;
			} else if (c == '+') {
				int i = rd(), j = rd();
				fu[i] = fu[j], p[i] = p[j];
			} else if (c == '-') {
				int i = rd(), j = rd();
				fu[i] = fu[j] ^ 1, p[i] = p[j];
			}
		}
     //vis 值 1表示 T 或 F，2表示在当前栈内，3表示为 U
		vis[n + 1] = vis[n + 2] = 1;
		vis[n + 3] = 3;
		for (int i = 1; i <= n; i ++ )
			if (p[i] == i && fu[i] == 1)
				vis[i] = 3, ans ++ ;
			else if (p[i] == i)
				vis[i] = 1;
		for (int i = 1; i <= n; i ++ ) {
			if (vis[i]) continue;
			int j = i;
			int cnt = 1;
			vector<int> g;
			g.pb(j);
			vis[j] = 2;
			while (vis[p[j]] == 0) {
				j = p[j];
				vis[j] = 2; 
				g.pb(j);
				cnt ++ ;
			}
			if (vis[p[j]] == 1) {
				for (int vv : g) vis[vv] = 1;
			} else if (vis[p[j]] == 3) {
				for (int vv : g) vis[vv] = 3;
				ans += cnt;
			} else {
				//int *ipx = 0; *ipx = 0; 
           //存在环
				for (int vv : g) vis[vv] = 1;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}