Editorial for Pondo LCA II

Use this editorial only when stuck, and do not copy-paste code from it.

Please be respectful to the problem author and editorialist. Submitting an official solution before solving the problem yourself is a bannable offence.

Approach

Root the tree at vertex $1$ and record an Euler tour of visits, storing the depth of the vertex at each tour index and the first index at which each vertex appears.

For a query $(u, v)$ , the lowest common ancestor is the vertex of minimum depth on the tour between $\mathrm{first}[u]$ and $\mathrm{first}[v]$ . With $n, q \le 10^5$ , a linear scan per query is too slow. Build a sparse table on the tour depths in $O(n \log n)$ , then each range minimum is $O(1)$ . Overall time is $O((n + q) \log n)$ to build and $O(n + q)$ to answer.

Solution (Python)

Code 1

import sys

sys.setrecursionlimit(1_000_000)


class SparseTable:
    def __init__(self, arr):
        self.arr = arr
        self.n = len(arr)
        self.log = [0] * (self.n + 1)
        self._compute_logs()
        self.st = self._build_sparse_table()

    def _compute_logs(self):
        for i in range(2, self.n + 1):
            self.log[i] = self.log[i // 2] + 1

    def _build_sparse_table(self):
        k = self.log[self.n] + 1
        st = [[(0, 0)] * k for _ in range(self.n)]
        for i in range(self.n):
            st[i][0] = (self.arr[i], i)
        j = 1
        while (1 << j) <= self.n:
            i = 0
            while i + (1 << j) - 1 < self.n:
                if st[i][j - 1][0] < st[i + (1 << (j - 1))][j - 1][0]:
                    st[i][j] = st[i][j - 1]
                else:
                    st[i][j] = st[i + (1 << (j - 1))][j - 1]
                i += 1
            j += 1
        return st

    def query(self, L, R):
        j = self.log[R - L + 1]
        left = self.st[L][j]
        right = self.st[R - (1 << j) + 1][j]
        if left[0] < right[0]:
            return left
        return right


data = list(map(int, sys.stdin.buffer.read().split()))
it = iter(data)
n = next(it)
q = next(it)
adj = [[] for _ in range(n)]
for _ in range(n - 1):
    u = next(it) - 1
    v = next(it) - 1
    adj[u].append(v)
    adj[v].append(u)

tour = []
tour_h = []
first = [-1] * n


def dfs(u, p, h):
    first[u] = len(tour)
    tour.append(u)
    tour_h.append(h)
    for v in adj[u]:
        if v == p:
            continue
        dfs(v, u, h + 1)
        tour.append(u)
        tour_h.append(h)


dfs(0, -1, 0)
st = SparseTable(tour_h)

out = []
for _ in range(q):
    u = next(it) - 1
    v = next(it) - 1
    l = first[u]
    r = first[v]
    if l > r:
        l, r = r, l
    _, idx = st.query(l, r)
    out.append(str(tour[idx] + 1))

print("\n".join(out))

Comments0

No comments yet

Be the first to comment.

Editorial for Pondo LCA II

Approach

Solution (Python)

Comments0

New comment