dp[u][0] := 考慮在 u 的子樹中，不選 u 的最大權獨立集
dp[u][1] := 考慮在 u 的子樹中，選 u 的最大權獨立集

int dp[N][2];

void dfs(int u, int p = -1) {
	dp[u][1] = w[u]; // weight of u
	for(int v:G[u]) if(v != p) {
    	dfs(v, u);
        dp[u][0] += max(dp[v][0], dp[v][1]);
        dp[u][1] += dp[v][0];
    }
}

int max_weight_indep_set() {
	dfs(1);
    return max(dp[1][0], dp[1][1]);
}

low[u] := 起點在 u 的子樹中的 back edges 的終點當中， dfs 序最小值

bool vis[N];
int tin[N], low[N], timer;
vi G[N];
void dfs(int u, int p = -1) {
    vis[u] = true;
    tin[u] = low[u] = timer++;
    for (int v : G[u]) if(v != p) {
        if (vis[v]) {
            low[u] = min(low[u], tin[v]); // 有一條從 u 開始的 back edge
        } else {
            dfs(v, u);
            low[u] = min(low[u], low[v]); // 轉移 dp
            if (low[v] == tin[v])
                IS_BRIDGE(u, v);
        }
    }
}

G is a connect graph.

* 8w Property
An edge (u, v) in G is a bridge iff u, v are not connected after removing (u, v) from G.
PROOF: trivial

* 8wcp Corollary
For any simple cycle C in a graph G, each edge in C is not a bridge in G.
PROOF: trivial

* 8wcpOrz Property
If G' is a connected subgraph of G, e is a bridge in G and e is in G',
 then e is also a bridge in G'.
PROOF: trivial

int tt, bcc_cnt, vis[N], low[N], bcc[N];
vi G[N], who[N], s;
void assign(int u, int bcc_id) {
	bcc[u] = bcc_id;
    who[bcc_id].PB(u);
}
void dfs(int u, int p = -1) {
	low[u] = vis[u] = ++tt;
	s.PB(u);
	for(int v:G[u]) if(v != p) {
		if(vis[v]) low[u] = min(low[u], vis[v]);
		else {
			dfs(v, u);
			if(low[v] > vis[u]) {
				++bcc_cnt;
				int k;
				do {
                	assign(k = s.back(), bcc_cnt);
					s.pop_back();
				} while(k != v);
			}
			low[u] = min(low[u], low[v]);
		}
	}
	if(pa == -1) {
		++bcc_cnt;
		for(; s.size(); s.pop_back())
			assign(s.back(), bcc_cnt);
	}
}

low[u] := 起點在 u 的子樹中的 back edges 的終點當中， dfs 序最小值

bool vis[N];
int tin[N], low[N], timer;
vi G[N];
void dfs(int u, int p = -1) {
    vis[u] = true;
    tin[u] = low[u] = timer++;
    int child_cnt = 0;
    for (int v : G[u]) if(v != p) {
        if (vis[v]) {
            low[u] = min(low[u], tin[v]); // 有一條從 u 開始的 back edge
        } else {
            dfs(v, u);
            if(p != -1 && low[v] >= tin[u])
            	IS_CUTPOINT(u); // u separates v from p
            child_cnt++, low[u] = min(low[u], low[v]); // 轉移 dp
        }
    }
    if(p == -1 && child_cnt > 1) IS_CUTPOINT(u);
}

G is a connected graph.

* 8e7 loves wiwi Lemma
If A, B are biconnected subgraphs of G and |V(A) ∩ V(B)| >= 2,
  then A U B is also a  biconnected subgraph.
PROOF. 我懶得用打的，現場講XD

* wiwi loves 8e7 Corollary
If A, B are biconnected subgraphs of G and |E(A) ∩ E(B)| >= 1,
  then A U B is also a  biconnected subgraph.
  
* 8w 1314 Observation
If (u, v) is an edge of G,
  then ({u, v}, {(u, v)}) is a biconnected subgraph of G.

* 8w forever Corollary
If e is an edge of G,
  then e belongs to at least one  biconnected subgraph of G.

int low[N], tin[N], id[N], t, cnt;
vi s, bcc[N], G[N];

void dfs(int u, int p = 0) {
	low[u] = tin[u] = ++t;
	s.PB(u);
	for(int v:G[u]) if(v != p) {
		if(tin[v]) low[u] = min(low[u], tin[v]);
		else {
			dfs(v, u);
			if(low[v] >= tin[u]) {
				cnt++;
				int k;
				do{
					k = s.back(); s.pop_back();
					bcc[k].PB(cnt);
				} while(k != v);
				bcc[u].PB(cnt);
			}
			low[u] = min(low[u], low[v]);
		}
	}
}

bool iscut[N];
vi T[N];

void build() {
	dfs(1);
	for(int i = 1; i <= n; i++) {
		if(bcc[i].size() > 1) {
			id[i] = ++cnt;
			iscut[id[i]] = true;
			for(int j:bcc[i]) {
				T[id[i]].PB(j);
				T[j].PB(id[i]);
			}
		} else {
			id[i] = bcc[i][0];
		}
	}
}

vi G[N], s;
int scc[N], d[N], low[N], cnt, tin;

void dfs(int u) {
	d[u] = low[u] = ++tin;
	s.PB(u);
	for(int v:G[u]) {
		if(!d[v]) {
			dfs(v);
			low[u] = min(low[u], low[v]);
		} else if(!scc[v]) {
			low[u] = min(low[u], d[v]);
		}
	}
	if(low[u] == d[u]) {
		cnt++;
		int t;
		do {
			t = s.back(); s.pop_back();
			scc[t] = cnt; 
		} while(t != u);
	}
}

V<pi> G[N];
vi rG[N], order;
int scc[N], vis[N];

void dfs(int u) {
	vis[u] = 1;
	for(int v:rG[u]) if(!vis[v])
		dfs(v);
	order.PB(u);
}

void dfs2(int u, int cnt) {
	scc[u] = cnt;
	d[u] = dis;
    for(int v:G[u]) {
		if(!scc[v]) dfs2(v, cnt);
	}
}

int main() {
	int n, m;
	cin >> n >> m;
	for(int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		G[u].EB(v);
		rG[v].EB(u);
	}
	for(int i = 1; i <= n; i++) if(!vis[i])
		dfs(i);
	reverse(ALL(order));
	int cnt = 0;
	for(int i:order) if(!scc[i]) {
		dfs2(i, ++cnt);
	}
}