我仿照题解2写的

#include<bits/stdc++.h>
using namespace std;
int n, m, s1, s2;
int dfn[2000005], low[2000005], tot, len;
vector<pair<int, int>> e[2000005];
vector<int> v;
vector<vector<int>> ans;
stack<int> st;
void tarjan(int u, int las) {
	dfn[u] = low[u] = ++tot;
	st.push(u);
	for(auto i : e[u]) {
		int v = i.first;
		if(i.second == (las ^ 1)) {
			continue;
		}
		if(dfn[v] == 0) {
			tarjan(v, i.second);
			low[u] = min(low[u], low[v]);
		} else {
			low[u] = min(low[u], dfn[v]);
		}
	}
	if(dfn[u] == low[u]){
		v.clear();
		v.push_back(u);
		while(st.top() != u){
			v.push_back(st.top());
			st.pop();
		}
		st.pop();
		ans.push_back(v);
	}
	return;
}
int main() {
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; i++) {
		scanf("%d %d", &s1, &s2);
		e[s1].push_back({s2, i << 1});
		e[s2].push_back({s1, i << 1 | 1});
	}
	for(int i = 1; i <= n; i++) {
		if(!dfn[i]) {
			tarjan(i, 0);
		}
	}
	printf("%d\n", (int)ans.size());
	for(auto i : ans){
		printf("%d ", (int)i.size());
		for(auto j : i){
			printf("%d ", j);
		}
		printf("\n");
	}
	return 0;
}

一个疑问：为什么tarjan函数写得如此简单（甚至比别的tarjan算法都简单）就能求出边双？