CF 1702G1 - Passable Paths (easy version)

Rating: 1900
Tags: dfs and similar, trees
Solve time: 2m 35s
Verified: yes

Solution

Problem Understanding

We are given a tree and several queries. Each query specifies a set of vertices. For every query, we must determine whether all selected vertices can lie on a single simple path of the tree.

A tree contains exactly one simple path between any two vertices. A query is considered passable if there exists some simple path such that every queried vertex belongs to that path. The path is allowed to contain additional vertices that were not part of the query.

The tree contains up to $2 \cdot 10^5$ vertices, while the total number of queried vertices across all queries is also at most $2 \cdot 10^5$. This immediately rules out any approach that repeatedly walks entire paths for every pair of queried vertices. An $O(n)$ operation per queried vertex would already become too expensive. We need preprocessing on the tree and then near-linear work in the size of each query.

Several situations are easy to misunderstand.

Consider a star-shaped tree:

Query:

{2,3,4}

The answer is NO.

Every pair of queried vertices has a path between them, but there is no single simple path containing all three leaves. Any path between two leaves passes through the center and reaches only those two leaves.

Another example:

1 - 2 - 3 - 4 - 5

Query:

{2,4}

The answer is YES.

The path does not need to start or end at queried vertices. The path from 1 to 5 contains both 2 and 4.

A more subtle case:

    2
    |
1 - 3 - 4 - 5

Query:

{1,2,5}

The answer is YES.

The path from 1 to 5 passes through vertex 3, but it does not pass through vertex 2. The actual path containing all queried vertices is 2-3-4-5, which also contains 1? No. So the answer is actually NO. This illustrates why checking only pairwise distances or pairwise connectivity is insufficient. We need to verify that every queried vertex belongs to one common path.

Approaches

A brute-force idea is to try all pairs of queried vertices as possible endpoints of the desired path. For each pair, construct their path and check whether every queried vertex lies on it.

In a tree, a vertex $x$ lies on the path between $a$ and $b$ exactly when

$$dist(a,x)+dist(x,b)=dist(a,b).$$

So we could test every endpoint pair and verify all queried vertices.

This is correct, but much too slow. If a query contains $k$ vertices, there are $O(k^2)$ endpoint pairs, and each pair requires checking $k$ vertices. That becomes $O(k^3)$. With $k$ reaching $2 \cdot 10^5$, this is hopeless.

The key observation comes from the geometry of trees.

Suppose all queried vertices lie on one simple path. Then among those vertices there must be two vertices that are the extremes of that path. Every other queried vertex lies on the unique path connecting those two extremes.

So instead of trying all endpoint pairs, we only need to find one candidate pair that could be the extremes.

A standard tree trick is that if we pick any queried vertex $s$, then the farthest queried vertex from $s$ is one endpoint of the diameter of the queried set. If we then start from that endpoint and again find the farthest queried vertex, we obtain the other endpoint.

Call these vertices $a$ and $b$.

If the queried vertices really lie on one path, then every queried vertex must belong to the path $a \leftrightarrow b$. Conversely, if every queried vertex belongs to that path, then the set is passable.

The remaining task is efficient distance computation. We preprocess the tree using DFS and binary lifting. Then every distance query can be answered in $O(\log n)$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(k^3)$ per query	$O(1)$	Too slow
Optimal	$O(n\log n + k\log n)$ total	$O(n\log n)$	Accepted

Algorithm Walkthrough

Preprocessing

Build the tree and run a DFS from the root.

Store the depth of every vertex and its binary lifting ancestors. This allows Lowest Common Ancestor queries in $O(\log n)$.

Using LCA, compute distances as:

$$dist(u,v)=depth[u]+depth[v]-2\cdot depth[lca(u,v)].$$

Query Processing

Take any queried vertex as the starting vertex $s$.
Compute the distance from $s$ to every queried vertex and find the farthest one. Call it $a$.

If the queried vertices lie on a path, one endpoint of their span must be farthest from any chosen vertex. 3. Compute the distance from $a$ to every queried vertex and find the farthest one. Call it $b$.

The pair $(a,b)$ is the diameter of the queried vertices. 4. For every queried vertex $v$, check whether

$$dist(a,v)+dist(v,b)=dist(a,b).$$ 5. If the equality holds for every queried vertex, output YES. 6. Otherwise output NO.

Why it works

Let $a$ and $b$ be the diameter endpoints among the queried vertices.

If all queried vertices lie on a single path, then that path must contain both diameter endpoints. Every queried vertex lies between them, so the distance equality holds for all vertices.

For the opposite direction, suppose every queried vertex satisfies

$$dist(a,v)+dist(v,b)=dist(a,b).$$

In a tree, this condition is true exactly when $v$ lies on the unique path from $a$ to $b$. Hence every queried vertex belongs to the same simple path $a \leftrightarrow b$. The set is passable.

The algorithm accepts exactly the passable sets.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())

g = [[] for _ in range(n + 1)]

for _ in range(n - 1):
    u, v = map(int, input().split())
    g[u].append(v)
    g[v].append(u)

LOG = (n + 1).bit_length()

up = [[0] * (n + 1) for _ in range(LOG)]
depth = [0] * (n + 1)

stack = [(1, 0)]

while stack:
    v, p = stack.pop()
    up[0][v] = p

    for to in g[v]:
        if to == p:
            continue
        depth[to] = depth[v] + 1
        stack.append((to, v))

for j in range(1, LOG):
    prev = up[j - 1]
    cur = up[j]
    for v in range(1, n + 1):
        cur[v] = prev[prev[v]]

def lca(a, b):
    if depth[a] < depth[b]:
        a, b = b, a

    diff = depth[a] - depth[b]

    bit = 0
    while diff:
        if diff & 1:
            a = up[bit][a]
        diff >>= 1
        bit += 1

    if a == b:
        return a

    for j in range(LOG - 1, -1, -1):
        if up[j][a] != up[j][b]:
            a = up[j][a]
            b = up[j][b]

    return up[0][a]

def dist(a, b):
    c = lca(a, b)
    return depth[a] + depth[b] - 2 * depth[c]

q = int(input())

ans = []

for _ in range(q):
    k = int(input())
    nodes = list(map(int, input().split()))

    start = nodes[0]

    a = start
    best = -1

    for v in nodes:
        d = dist(start, v)
        if d > best:
            best = d
            a = v

    b = a
    best = -1

    for v in nodes:
        d = dist(a, v)
        if d > best:
            best = d
            b = v

    diameter = dist(a, b)

    ok = True

    for v in nodes:
        if dist(a, v) + dist(v, b) != diameter:
            ok = False
            break

    ans.append("YES" if ok else "NO")

sys.stdout.write("\n".join(ans))

After preprocessing, every distance query requires only one LCA computation. For each query we perform two scans to find the diameter endpoints and one final scan for verification.

The DFS is implemented iteratively to avoid Python recursion depth issues on a tree with $2 \cdot 10^5$ vertices.

A common mistake is to assume the diameter endpoints of the entire tree matter. We only care about the diameter among the queried vertices. The two farthest-point searches are performed exclusively on the vertices appearing in the query.

Another subtle point is the path-membership test. Checking only that a vertex is close to the diameter path is not enough. The exact equality

$$dist(a,v)+dist(v,b)=dist(a,b)$$

must hold.

Worked Examples

Sample 1, Query `{3,2,5}`

Tree:

1-2
2-3
2-4
4-5

Step	Value
Start vertex	3
Farthest from 3	5
a	5
Farthest from 5	3
b	3
dist(a,b)	3

Verification:

Vertex	dist(5,v)	dist(v,3)	Sum	On path?
3	3	0	3	Yes
2	2	1	3	Yes
5	0	3	3	Yes

Answer: YES.

Every queried vertex lies on the path 5-4-2-3.

Sample 1, Query `{1,2,3,4,5}`

Step	Value
Start vertex	1
Farthest from 1	5
a	5
Farthest from 5	1
b	1
dist(a,b)	3

Verification:

Vertex	dist(5,v)+dist(v,1)	Equals 3?
1	3	Yes
2	3	Yes
3	5	No
4	3	Yes
5	3	Yes

Since vertex 3 fails, the answer is NO.

This query contains vertices from two different branches of the tree, so no single path can contain them all.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n\log n + K\log n)$	DFS and binary lifting preprocessing plus all distance computations, where $K$ is the total number of queried vertices
Space	$O(n\log n)$	Ancestor table and tree storage

The total queried size is at most $2 \cdot 10^5$, so the overall complexity is comfortably within the limits. Binary lifting keeps every distance query logarithmic.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)

    input = sys.stdin.readline

    n = int(input())
    g = [[] for _ in range(n + 1)]

    for _ in range(n - 1):
        u, v = map(int, input().split())
        g[u].append(v)
        g[v].append(u)

    LOG = (n + 1).bit_length()

    up = [[0] * (n + 1) for _ in range(LOG)]
    depth = [0] * (n + 1)

    stack = [(1, 0)]

    while stack:
        v, p = stack.pop()
        up[0][v] = p

        for to in g[v]:
            if to != p:
                depth[to] = depth[v] + 1
                stack.append((to, v))

    for j in range(1, LOG):
        for v in range(1, n + 1):
            up[j][v] = up[j - 1][up[j - 1][v]]

    def lca(a, b):
        if depth[a] < depth[b]:
            a, b = b, a

        diff = depth[a] - depth[b]

        bit = 0
        while diff:
            if diff & 1:
                a = up[bit][a]
            diff >>= 1
            bit += 1

        if a == b:
            return a

        for j in range(LOG - 1, -1, -1):
            if up[j][a] != up[j][b]:
                a = up[j][a]
                b = up[j][b]

        return up[0][a]

    def dist(a, b):
        c = lca(a, b)
        return depth[a] + depth[b] - 2 * depth[c]

    q = int(input())
    out = []

    for _ in range(q):
        k = int(input())
        nodes = list(map(int, input().split()))

        start = nodes[0]

        a = max(nodes, key=lambda x: dist(start, x))
        b = max(nodes, key=lambda x: dist(a, x))

        dab = dist(a, b)

        ok = all(dist(a, v) + dist(v, b) == dab for v in nodes)

        out.append("YES" if ok else "NO")

    return "\n".join(out)

# provided sample
assert run(
"""5
1 2
2 3
2 4
4 5
5
3
3 2 5
5
1 2 3 4 5
2
1 4
3
1 3 5
3
1 5 4
"""
) == "YES\nNO\nYES\nNO\nYES"

# single vertex
assert run(
"""1
1
1
1
"""
) == "YES"

# chain tree, all vertices on one path
assert run(
"""5
1 2
2 3
3 4
4 5
1
3
1 3 5
"""
) == "YES"

# star tree, multiple leaves
assert run(
"""5
1 2
1 3
1 4
1 5
1
3
2 3 4
"""
) == "NO"

# path with interior vertices
assert run(
"""5
1 2
2 3
3 4
4 5
1
2
2 4
"""
) == "YES"

Test input	Expected output	What it validates
Single vertex tree	YES	Minimum input size
Chain tree	YES	Entire query already lies on one path
Star tree	NO	Multiple branches cannot fit on one path
Two interior vertices on chain	YES	Endpoints of the path need not belong to the query

Edge Cases

Consider the star:

The diameter endpoints become two leaves, for example 2 and 3. Vertex 4 fails the equality check because it is not on the path 2-1-3. The algorithm outputs NO.

Consider a single queried vertex:

The diameter endpoints are the same vertex. The distance is zero, and the verification succeeds immediately. The algorithm outputs YES.

Consider a path where queried vertices are not endpoints:

1-2-3-4-5
query = {2,4}

The algorithm chooses 2 and 4 as diameter endpoints. Both satisfy the path condition, so the answer is YES. This confirms that the path may extend beyond queried vertices.

Consider a branching tree:

Query:

{1,2,5}

The diameter endpoints are 1 and 2. Vertex 5 does not satisfy the equality for path 1-3-2. The algorithm correctly outputs NO, detecting that the queried vertices occupy different branches.

CF 1702G1 - Passable Paths (easy version)

Solution

Problem Understanding

Approaches

Algorithm Walkthrough

Preprocessing

Query Processing

Why it works

Python Solution

Worked Examples

Sample 1, Query {3,2,5}

Sample 1, Query {1,2,3,4,5}

Complexity Analysis

Test Cases

Edge Cases

Sample 1, Query `{3,2,5}`

Sample 1, Query `{1,2,3,4,5}`