最近公共祖先（LCA）

luogu 模板：https://www.luogu.com.cn/problem/P3379

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
int n, m, root, d[N], p[N][30], lg[N];
vector <int> g[N];
void dfs(int u, int fa){
	p[u][0] = fa;
	d[u] = d[fa] + 1;
	for (int i = 1; i <= lg[d[u]]; i ++ )
		p[u][i] = p[p[u][i - 1]][i - 1];
		// u 的 2^i 的祖先等于 u 的 2^(i-1) 的祖先的 2^(i-1) 的祖先
	for (auto v : g[u])
		if (v != fa)
			dfs(v, u);
}
int lca(int x, int y){
	if(d[x] < d[y]) swap(x, y);
	while (d[x] > d[y])
		x = p[x][lg[d[x] - d[y]] - 1];
	if (x == y) return x;
	for (int k = lg[d[x]] - 1; k >= 0; k -- )
		if (p[x][k] != p[y][k]){
			x = p[x][k];
			y = p[y][k];
		}
	return p[x][0];
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin >> n >> m >> root;
	for (int i = 1; i < n; i ++ ){
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for (int i = 1; i <= n; i ++ )	//预处理 log
		lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
	dfs(root, 0);	//找到每个点的祖先 
	for (int i = 1; i <= m; i ++ ){
		int x, y;
		cin >> x >> y;
		cout << lca(x, y) << "\n";
	}
	return 0;
}

标签：lg,return,fa,祖先,luogu,LCA,cin,int,公共
来源： https://www.cnblogs.com/Hamine/p/16191722.html