LCA之倍增法

//#include <bits/stdc++.h>
#include<iostream>
#include<vector>
using namespace std;
const int N = 5e5 + 20;
int far[N][20], dep[N];
vector<int>g[N];
int n, m, s, u, v;
void dfs(int u, int father) {
	dep[u] = dep[father] + 1;
	far[u][0] = father;
	for (int i = 1; i <= 19; i++) {
		far[u][i] = far[far[u][i - 1]][i - 1];
	}
	for (int v : g[u]) {
		if (v != father) dfs(v, u);
	}
}
int lca(int u, int v) {
	if (dep[u] < dep[v])swap(u, v);
	for (int i = 19; i >= 0; i--) {
		if (dep[far[u][i]] >= dep[v])
			u = far[u][i];
	}
	if (u == v)return v;
	for (int i = 19; i >= 0; i--) {
		if (far[u][i] != far[v][i])
			u = far[u][i], v = far[v][i];
	}
	return far[u][0];
}
int main() {
	cin >> n >> m >> s;
	for (int i = 1; i <= n - 1; i++) {
		scanf("%d %d", &u, &v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(s, 0);
	for (int i = 1; i <= m; i++) {
		scanf("%d %d", &u, &v);
		cout << lca(u, v) << endl;
	}
	return 0;
}

// 运行更快
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 5e5 + 10, M = 2 * N;
int n, m, s, a, b;
int dep[N], fa[N][22];
int h[N], to[M], ne[M], tot;
void add(int a, int b) {
    to[++tot] = b, ne[tot] = h[a], h[a] = tot;
}
void dfs(int x, int f) {
    dep[x] = dep[f] + 1; fa[x][0] = f;
    for (int i = 0; i <= 20; i++)
        fa[x][i + 1] = fa[fa[x][i]][i];
    for (int i = h[x]; i; i = ne[i])
        if (to[i] != f) dfs(to[i], x);
}
int lca(int x, int y) {
    if (dep[x] < dep[y]) swap(x, y);
    for (int i = 20; ~i; i--)
        if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];
    if (x == y) return y;

    for (int i = 20; ~i; i--)
        if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}
int main() {
    scanf("%d%d%d", &n, &m, &s);
    for (int i = 1; i < n; i++) {
        scanf("%d%d", &a, &b);
        add(a, b); add(b, a);
    }
    dfs(s, 0);
    while (m--) {
        scanf("%d%d", &a, &b);
        printf("%d\n", lca(a, b));
    }
    return 0;
}