Tree Queries (Codeforces 1328E)
Tags: Others, trees

You are given a rooted tree consisting of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. The root of the tree is a vertex number $$$1$$$.A tree is a connected undirected graph with $$$n-1$$$ edges.You are given $$$m$$$ queries. The $$$i$$$-th query consists of the set of $$$k_i$$$ distinct vertices $$$v_i[1], v_i[2], \dots, v_i[k_i]$$$. Your task is to say if there is a path from the root to some vertex $$$u$$$ such that each of the given $$$k$$$ vertices is either belongs to this path or has the distance $$$1$$$ to some vertex of this path.

Input: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le m \le 2 \cdot 10^5$$$) — the number of vertices in the tree and the number of queries.Each of the next $$$n-1$$$ lines describes an edge of the tree. Edge $$$i$$$ is denoted by two integers $$$u_i$$$ and $$$v_i$$$, the labels of vertices it connects $$$(1 \le u_i, v_i \le n, u_i \ne v_i$$$).It is guaranteed that the given edges form a tree.The next $$$m$$$ lines describe queries. The $$$i$$$-th line describes the $$$i$$$-th query and starts with the integer $$$k_i$$$ ($$$1 \le k_i \le n$$$) — the number of vertices in the current query. Then $$$k_i$$$ integers follow: $$$v_i[1], v_i[2], \dots, v_i[k_i]$$$ ($$$1 \le v_i[j] \le n$$$), where $$$v_i[j]$$$ is the $$$j$$$-th vertex of the $$$i$$$-th query.It is guaranteed that all vertices in a single query are distinct.It is guaranteed that the sum of $$$k_i$$$ does not exceed $$$2 \cdot 10^5$$$ ($$$\sum\limits_{i=1}^{m} k_i \le 2 \cdot 10^5$$$).

Output: For each query, print the answer — "YES", if there is a path from the root to some vertex $$$u$$$ such that each of the given $$$k$$$ vertices is either belongs to this path or has the distance $$$1$$$ to some vertex of this path and "NO" otherwise.

Note: The picture corresponding to the example:Consider the queries.The first query is $$$[3, 8, 9, 10]$$$. The answer is "YES" as you can choose the path from the root $$$1$$$ to the vertex $$$u=10$$$. Then vertices $$$[3, 9, 10]$$$ belong to the path from $$$1$$$ to $$$10$$$ and the vertex $$$8$$$ has distance $$$1$$$ to the vertex $$$7$$$ which also belongs to this path.The second query is $$$[2, 4, 6]$$$. The answer is "YES" as you can choose the path to the vertex $$$u=2$$$. Then the vertex $$$4$$$ has distance $$$1$$$ to the vertex $$$1$$$ which belongs to this path and the vertex $$$6$$$ has distance $$$1$$$ to the vertex $$$2$$$ which belongs to this path.The third query is $$$[2, 1, 5]$$$. The answer is "YES" as you can choose the path to the vertex $$$u=5$$$ and all vertices of the query belong to this path.The fourth query is $$$[4, 8, 2]$$$. The answer is "YES" as you can choose the path to the vertex $$$u=9$$$ so vertices $$$2$$$ and $$$4$$$ both have distance $$$1$$$ to the vertex $$$1$$$ which belongs to this path and the vertex $$$8$$$ has distance $$$1$$$ to the vertex $$$7$$$ which belongs to this path.The fifth and the sixth queries both have answer "NO" because you cannot choose suitable vertex $$$u$$$.

Sample Inputs:
10 6
1 2
1 3
1 4
2 5
2 6
3 7
7 8
7 9
9 10
4 3 8 9 10
3 2 4 6
3 2 1 5
3 4 8 2
2 6 10
3 5 4 7

Sample Outputs:
YES
YES
YES
YES
NO
NO

Solution (Python):
import sys
input_func = sys.stdin.readline
(num_nodes, num_queries) = map(int, input_func().split())
adj_list = [[] for _ in range(num_nodes + 1)]
for _ in range(num_nodes - 1):
    (u, v) = map(int, input_func().split())
    adj_list[u].append(v)
    adj_list[v].append(u)
parent_dict = {1: 1}
in_time_array = [0] * (num_nodes + 1)
out_time_array = [0] * (num_nodes + 1)
stack = [(1, 0)]
count = 0
while stack:
    (node, visited) = stack.pop()
    if not visited:
        count += 1
        in_time_array[node] = count
        stack.append((node, 1))
        for neighbor in adj_list[node]:
            if neighbor not in parent_dict:
                parent_dict[neighbor] = node
                stack.append((neighbor, 0))
    else:
        out_time_array[node] = count
for _ in range(num_queries):
    query_list = list(map(int, input_func().split()))[1:]
    (low, high) = (1, num_nodes)
    for node_query in query_list:
        low = max(low, in_time_array[parent_dict[node_query]])
        high = min(high, out_time_array[parent_dict[node_query]])
    print('YES' if low <= high else 'NO')

Solution (Python):
import sys
input = sys.stdin.readline
(n, q) = map(int, input().split())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
    (u, v) = map(int, input().split())
    adj[u].append(v)
    adj[v].append(u)
parent = {1: 1}
in_time = [0] * (n + 1)
out_time = [0] * (n + 1)
stack = [(1, 0)]
cnt = 0
while stack:
    (node, visited) = stack.pop()
    if not visited:
        cnt += 1
        in_time[node] = cnt
        stack.append((node, 1))
        for v in adj[node]:
            if v not in parent:
                parent[v] = node
                stack.append((v, 0))
    else:
        out_time[node] = cnt
for i in range(q):
    query = list(map(int, input().split()))[1:]
    (low, high) = (1, n)
    for i in query:
        (low, high) = (max(low, in_time[parent[i]]), min(high, out_time[parent[i]]))
    print('YES' if low <= high else 'NO')

Solution (Python):
import sys
input = sys.stdin.readline
(n, q) = map(int, input().split())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
    (u, v) = map(int, input().split())
    adj[u].append(v)
    adj[v].append(u)
parent = {1: 1}
in_time = [0] * (n + 1)
out_time = [0] * (n + 1)
stack = [(1, 0)]
cnt = 0
while stack:
    (node, visited) = stack.pop()
    if not visited:
        cnt += 1
        in_time[node] = cnt
        stack.append((node, 1))
        for v in adj[node]:
            if v not in parent:
                parent[v] = node
                stack.append((v, 0))
    else:
        out_time[node] = cnt
for i in range(q):
    query = list(map(int, input().split()))[1:]
    (low, high) = (1, n)
    for i in query:
        (low, high) = (max(low, in_time[parent[i]]), min(high, out_time[parent[i]]))
    print('YES' if low <= high else 'NO')