强连通分量

强连通分量有两种解法
1.一种是tarjan算法（不得不说这人很强）
2.两边dfs的算法

//1.tarjan算法
#include<iostream>
using namespace std;
#include<stack>
#include<vector>
const int maxn = 1e5+5;
bool inst[maxn];
int cnt = 0,scc[maxn];
int dfn[maxn],low[maxn];
stack<int> st;
vector<int> son[maxn];
int tt = 0;
void dfs(int u){
	dfn[u] = low[u] = ++cnt;
	st.push(u);inst[u] = 1;
	for(int i=0;i<son[u].size();i++){
		int v = son[u][i];
		if(!dfn[v]){
			dfs(v);
			low[u] = min(low[u],low[v]);
		}
		else if(inst[v]){
			low[u] = min(low[u],dfn[v]);
		}
	}
	if(low[u] == dfn[u]){
		++tt;
		while(1){
			int v = st.top();
			inst[v] = 0;
			scc[v] = tt;
			st.pop();
			if(u == v){
				break;
			}
		}
	}
}
int main(){
	int n,m;
	cin >> n >> m;
	for(int i=1;i<=m;i++){
		int u,v;
		cin >> u >> v;
		son[u].push_back(v);
	}
	dfs(1);
	for(int i=1;i<=tt;i++){
		for(int j=1;j<=n;j++){
			if(scc[j] == i){
				cout << j <<"  ";
			}
		}
		cout << endl;
	}
}


2.两遍dfs
#include<iostream>
using namespace std;
#include<vector>
const int maxn = 1e5 + 5;
vector<int> s;
vector<int> son1[maxn];
vector<int> son2[maxn];
int vis[maxn], color[maxn];
int cnt = 0;
void dfs1(int u) {
	vis[u] = 1;
	for (int i = 0; i < son1[u].size(); i++) {
		int v = son1[u][i];
		if (!vis[v])
			dfs1(v);
	}
	s.push_back(u);
}
void dfs2(int u) {
	color[u] = cnt;
	for (int i = 0; i < son2[u].size(); i++) {
		int v = son2[u][i];
		if (!color[v]) {
			dfs2(v);
		}
	}
}
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int u, v;
		cin >> u >> v;
		son1[u].push_back(v);
		son2[v].push_back(u);
	}
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {
			dfs1(i);
		}
	}
	for (int i = n-1; i >= 0; i--) {
		if (!color[s[i]]) {
			++cnt;
			dfs2(s[i]);
		}
	}
	for(int i=1;i<=cnt;i++){
		for(int j=1;j<=n;j++){
			if(color[j] == i){
				cout << j << "  ";
			}
		}
		cout << endl;
	}
}

个人感觉两遍dfs的好写点