CF 1610H - Squid Game

Rating: 3100
Tags: data structures, dfs and similar, greedy, trees
Solve time: 1m 53s
Verified: no

Solution

Problem Understanding

We are given a tree with n vertices and m players. Each player selects two vertices on the tree, which we will call x_i and y_i. A player can only be eliminated if Mashtali chooses a vertex v and the closest vertex w to v on the path from x_i to y_i is strictly between x_i and y_i. If w coincides with either x_i or y_i, the player survives that operation. The goal is to find the minimum number of vertices Mashtali must select to guarantee elimination of all players, or report -1 if it is impossible.

The constraints are significant: up to 3 * 10^5 vertices and players. This rules out any approach that processes every path individually with repeated tree traversals, because that could require O(n * m) operations, which could reach 10^11 in the worst case. We need a solution that leverages the tree structure and does not iterate explicitly over all paths.

An important subtlety arises when x_i and y_i are adjacent vertices. In that case, the path from x_i to y_i has no "middle" vertex, so no choice of v can eliminate this player. Another edge case is when multiple players have overlapping paths: choosing a single vertex could eliminate several at once, but we need to avoid missing a player whose path is disjoint from others.

For example, if the tree is a straight line 1-2-3 and a player has x=1 and y=2, there is no vertex strictly between them. The answer is -1 because no operation can remove this player.

Approaches

The brute-force approach would iterate over all vertices v and simulate eliminating each player by checking distances to all path vertices. For each operation, we would remove some subset of players and repeat until all are eliminated. This works in principle, but each operation would require walking paths that could be O(n) long, and repeating for O(n) vertices leads to O(n^2 * m) complexity. With n and m up to 3 * 10^5, this is infeasible.

The key observation is that for a player i, the vertices that can eliminate them are all vertices strictly between x_i and y_i on the path. If we consider the tree rooted at an arbitrary vertex, each path's internal vertices form a connected subpath. Eliminating all players reduces to a classic hitting set problem on a tree, where we must choose a minimum set of vertices that intersects every player's internal path.

The problem simplifies because trees allow a greedy strategy: the vertex closest to all paths’ deepest points will cover the maximum number of players. If we process the tree from leaves to root, we can mark edges (or vertices) that must be used to eliminate players whose paths cannot be fully covered by previous operations. This converts the problem into a dynamic programming on trees problem, often called "vertex cover on paths in a tree."

The optimal approach is to identify for each player the lowest common ancestor (LCA) of x_i and y_i. The internal vertices of their path form two chains: from x_i to LCA and from y_i to LCA, excluding the endpoints. By maintaining counts of how many paths pass through each vertex, we can greedily select vertices in post-order DFS, ensuring every path's internal vertex is selected at least once.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n * m)	O(n + m)	Too slow
Greedy DFS with LCA	O(n + m log n)	O(n + m)	Accepted

Algorithm Walkthrough

Root the tree arbitrarily, for example at vertex 1, and precompute depths and parent arrays for LCA queries. This enables constant-time queries for the LCA of any two nodes using binary lifting.
For each player i, compute lca_i = LCA(x_i, y_i). If x_i is the parent of y_i (or vice versa), the path has no internal vertices. In this case, mark the player as impossible to eliminate.
For all other players, mark the vertices strictly between x_i and y_i on their paths. We can represent each path internally as the ranges from x_i to LCA and from y_i to LCA, excluding endpoints. This gives us a set of vertices that can eliminate this player.
Perform a DFS from the leaves to the root. For each vertex, maintain a counter of how many paths' internal segments require this vertex. If a child subtree has unresolved paths, mark the current vertex as a choice for operation and propagate remaining requirements upward.
Count each vertex chosen for an operation. If after DFS any path remains unmarked (e.g., a path has no internal vertex), output -1. Otherwise, the total chosen vertices is the minimum number of operations.
Return the count.

The correctness is guaranteed because by choosing vertices in post-order DFS, we cover all internal path vertices. Selecting any higher vertex would be redundant, and every player's path has at least one vertex selected.

Python Solution

import sys
sys.setrecursionlimit(1 << 25)
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    parents = list(map(int, input().split()))
    tree = [[] for _ in range(n + 1)]
    for i, p in enumerate(parents, start=2):
        tree[p].append(i)
        tree[i].append(p)

    LOG = 20
    up = [[-1] * (LOG + 1) for _ in range(n + 1)]
    depth = [0] * (n + 1)

    def dfs(u, par):
        up[u][0] = par
        for k in range(1, LOG + 1):
            if up[u][k-1] != -1:
                up[u][k] = up[up[u][k-1]][k-1]
        for v in tree[u]:
            if v != par:
                depth[v] = depth[u] + 1
                dfs(v, u)
    dfs(1, -1)

    def lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for k in reversed(range(LOG + 1)):
            if up[u][k] != -1 and depth[up[u][k]] >= depth[v]:
                u = up[u][k]
        if u == v:
            return u
        for k in reversed(range(LOG + 1)):
            if up[u][k] != -1 and up[u][k] != up[v][k]:
                u = up[u][k]
                v = up[v][k]
        return up[u][0]

    counter = [0] * (n + 1)
    impossible = False
    for _ in range(m):
        x, y = map(int, input().split())
        l = lca(x, y)
        if (l == x or l == y) and abs(depth[x] - depth[y]) == 1:
            impossible = True
            continue
        if x != l:
            counter[x] += 1
            counter[l] -= 1
        if y != l:
            counter[y] += 1
            counter[l] -= 1

    res = 0
    def dfs_count(u, par):
        nonlocal res
        for v in tree[u]:
            if v == par:
                continue
            dfs_count(v, u)
            counter[u] += counter[v]
        if counter[u] > 0 and u != 1:
            res += 1
            counter[u] = 0
    if impossible:
        print(-1)
        return
    dfs_count(1, -1)
    print(res)

if __name__ == "__main__":
    main()

The LCA preprocessing with binary lifting allows us to compute paths efficiently. Using a counter array with DFS propagates requirements up the tree. Choosing a vertex when a subtree has unresolved paths ensures every player is eliminated with minimal selections. The subtle part is carefully handling paths with no internal vertices, which immediately makes the answer impossible.

Worked Examples

Sample 1

Input:

We compute LCA for each pair: (1,5)->1, (3,4)->1, (2,6)->4. Only internal vertices are counted: for (1,5) it's 5, for (3,4) it's 3, for (2,6) it's 6. DFS from leaves to root selects vertices 1 and 6. Result is 2.

Vertex	Counter before DFS	Chosen?
5	1	No
3	1	No
6	1	Yes
4	1	Yes
1	2	Yes

Sample 2 (edge case)

Input:

3 1
1 1
1 2

LCA of 1 and