2019 CCPC 重现赛 1006 基环树

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
const int mod = 998244353;
typedef long long ll;
vector<int> g[maxn];
int t,n,m,tot = 0,sz = 0,cnt;
int dfn[maxn],f[maxn];
vector<int> pot[maxn];
void dfs(int u,int fa) {
	dfn[u] = ++sz;
	bool t = false;
	for(int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if(v == fa) continue;
		if(!dfn[v]) {
			f[v] = u;
			dfs(v,u);
		} else if(dfn[v] < dfn[u]) continue;
		else if(dfn[v] > dfn[u]) {
			++cnt;
			for(int p = v; p != u; p = f[p])
				pot[cnt].push_back(p);
			pot[cnt].push_back(u);
			continue;
		}
	}
}
ll fpow(ll a,ll b) {
	ll r = 1;
	while(b) {
		if(b & 1) r = r * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return r;
}
int main() {
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; i++) {
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	ll ans = 1;
	for(int i = 1; i <= n; i++) {
		if(!dfn[i]) {
			dfs(i,0);
		}
	}
	for(int i = 1; i <= cnt; i++) {
		ans = ans * (fpow(2,pot[i].size()) - 1) % mod;
		m -= pot[i].size();
	}
	ans = ans * fpow(2,m) % mod;
	printf("%lld\n",ans);
	return 0;
}