倍增算法

预处理 O(nlogn)

单次询问 O(logn)

void dfs(int u, int fa)
{
	for (int i = hd[u]; i; i = g[i].nxt)
	{
		int v = g[i].to;
		if (v == fa)
			continue;
		d[v] = d[u] + 1;
		anc[v][0] = u;
		dfs(v, u);
	}
	return;
}
void init()
{
	for (int j = 1; j <= 18; j++)
		for (int i = 1; i <= n; i++)
			anc[i][j] = anc[anc[i][j - 1]][j - 1];
}
int LCA(int u, int v)
{
	if (d[u] < d[v])
		swap(u, v);
	for (int i = 18; i >= 0; i--)
		if (d[anc[u][i]] >= d[v])
			u = anc[u][i];
	if (u == v)
		return u;
	for (int i = 18; i >= 0; i--)
		if (anc[u][i] != anc[v][i])
			u = anc[u][i], v = anc[v][i];
	return anc[u][0];
}

Tarjan算法（离线）

时间复杂度 O(n + m)

inline int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
void dfs(int u)
{
	fa[u] = u;
	vis[u] = 1;
	for (int i = 0; i < G[u].size(); i++)
	{
		int v = G[u][i];
		if (vis[v])
			continue;
		dfs(v);
		fa[v] = u;
	}
	for (int i = 0; i < Q[u].size(); i++)
	{
		int v = Q[u][i].first;
		if (!vis[v])
			continue;
		ans[Q[u][i].second] = find(v);
	}
}