Sobre

Assume que um vertice eh ancestral dele mesmo, ou seja,

se a eh ancestral de b, lca(a, b) = a

dist(a, b) retorna a distancia entre a e b

Complexidades:

build - O(n)

lca - O(1)

dist - O(1)

Link original: lca.cpp

Código

  template<typename T> struct rmq {
	vector<T> v;
	int n; static const int b = 30;
	vector<int> mask, t;

	int op(int x, int y) { return v[x] < v[y] ? x : y; }
	int msb(int x) { return __builtin_clz(1)-__builtin_clz(x); }
	rmq() {}
	rmq(const vector<T>& v_) : v(v_), n(v.size()), mask(n), t(n) {
		for (int i = 0, at = 0; i < n; mask[i++] = at |= 1) {
			at = (at<<1)&((1<<b)-1);
			while (at and op(i, i-msb(at&-at)) == i) at ^= at&-at;
		}
		for (int i = 0; i < n/b; i++) t[i] = b*i+b-1-msb(mask[b*i+b-1]);
		for (int j = 1; (1<<j) <= n/b; j++) for (int i = 0; i+(1<<j) <= n/b; i++)
			t[n/b*j+i] = op(t[n/b*(j-1)+i], t[n/b*(j-1)+i+(1<<(j-1))]);
	}
	int small(int r, int sz = b) { return r-msb(mask[r]&((1<<sz)-1)); }
	T query(int l, int r) {
		if (r-l+1 <= b) return small(r, r-l+1);
		int ans = op(small(l+b-1), small(r));
		int x = l/b+1, y = r/b-1;
		if (x <= y) {
			int j = msb(y-x+1);
			ans = op(ans, op(t[n/b*j+x], t[n/b*j+y-(1<<j)+1]));
		}
		return ans;
	}
};

namespace lca {
	vector<int> g[MAX];
	int v[2*MAX], pos[MAX], dep[2*MAX];
	int t;
	rmq<int> RMQ;

	void dfs(int i, int d = 0, int p = -1) {
		v[t] = i, pos[i] = t, dep[t++] = d;
		for (int j : g[i]) if (j != p) {
			dfs(j, d+1, i);
			v[t] = i, dep[t++] = d;
		}
	}
	void build(int n, int root) {
		t = 0;
		dfs(root);
		RMQ = rmq<int>(vector<int>(dep, dep+2*n-1));
	}
	int lca(int a, int b) {
		a = pos[a], b = pos[b];
		return v[RMQ.query(min(a, b), max(a, b))];
	}
	int dist(int a, int b) {
		return dep[pos[a]] + dep[pos[b]] - 2*dep[pos[lca(a, b)]];
	}
}