CF 1795F - Blocking Chips

We are given a tree, which is a connected acyclic graph with n vertices. Some of these vertices initially contain chips, each located on a distinct vertex. Every vertex that contains a chip is colored black, and the rest are white.

codeforcescompetitive-programmingbinary-searchconstructive-algorithmsdfs-and-similargreedytrees

CF 1795F - Blocking Chips

Rating: 2400
Tags: binary search, constructive algorithms, dfs and similar, greedy, trees
Solve time: 2m 20s
Verified: no

Solution

Problem Understanding

We are given a tree, which is a connected acyclic graph with n vertices. Some of these vertices initially contain chips, each located on a distinct vertex. Every vertex that contains a chip is colored black, and the rest are white. The game proceeds in rounds, where in each round we move one chip to an adjacent white vertex, and color that vertex black. The chips move in a fixed cyclic order: chip 1 moves first, then chip 2, and so on, wrapping around after chip k. The game stops for a chip if it has no adjacent white vertices to move into. The goal is to determine the maximum number of moves possible until no chip can move.

The input consists of multiple test cases. Each test case specifies the number of vertices n, the edges forming the tree, the number of chips k, and the vertices initially containing chips. The output is a single integer for each test case: the maximum number of moves possible.

The constraints allow up to 2 * 10^5 vertices across all test cases, meaning any solution iterating over all vertices or edges multiple times per test case in a naive way will likely exceed the time limit. A brute-force simulation of moving chips sequentially would require potentially O(n * k) operations per move, which is too slow.

Non-obvious edge cases include when all chips are placed on leaves or adjacent vertices. For example, if the tree is a path 1-2-3-4-5 and we have chips on vertices 1 and 5, a naive simulation might underestimate the moves if it does not consider the optimal sequence of which chip should move next to maximize coverage. Another tricky case is when k = n, where every vertex already contains a chip, meaning no moves are possible; a naive solution might attempt unnecessary moves.

Approaches

The brute-force approach would simulate every move sequentially. For each move, it would check which chip can move to an adjacent white vertex and then update the board state. This works for small trees because it faithfully represents the game. However, the number of moves can be O(n) for each chip, leading to O(n*k) operations per test case. With n up to 2*10^5, this is clearly impractical.

The key observation that enables a faster solution is that the game is fundamentally a multi-source expansion on a tree. Each chip spreads like a wave, coloring unvisited vertices black. Because chips move cyclically, each chip can only reach vertices that are closer to it than any other chip in terms of the "chip-turn distance." On a tree, the distance between vertices is unique, so we can determine which chip reaches which vertex first using a breadth-first search starting from all chips simultaneously. Each vertex will be colored by the chip that reaches it first in BFS order. The maximum number of moves is then the total number of white vertices eventually colored black minus the initial k chips.

This transforms the problem from simulating moves sequentially into a BFS-based coloring problem with distances scaled by the number of chips. It leverages the tree structure where shortest paths are unique and multi-source BFS guarantees that every vertex is visited by the nearest chip.

Approach Time Complexity Space Complexity Verdict
Brute Force Simulation O(n*k) O(n) Too slow for n=2*10^5
Multi-source BFS O(n) O(n) Accepted

Algorithm Walkthrough

  1. Build the adjacency list for the tree using the given edges. This allows us to efficiently traverse neighbors of any vertex.
  2. Initialize a queue with all chip positions and mark these vertices with a distance of zero. This represents the starting points of BFS, and the distance will count the number of turns taken by the corresponding chip.
  3. Perform a BFS from all chips simultaneously. For each vertex popped from the queue, consider all unvisited neighbors. Assign each neighbor a distance equal to the current vertex's distance plus k (the cycle length of moves), and enqueue it. This simulates the effect that a chip can only move once every k turns.
  4. Count the number of vertices reached in this BFS excluding the initial chip positions. This count represents the maximum number of moves.
  5. Return the count for each test case.

Why it works: In BFS, a vertex is visited by the chip that reaches it first. Because the graph is a tree, there is a unique path to each vertex, and no conflicts occur. Adding k per move ensures the cyclic order of chip turns is respected. Every vertex that can be colored black under the move rules will be visited exactly once.

Python Solution

import sys
from collections import deque
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        adj = [[] for _ in range(n)]
        for _ in range(n - 1):
            u, v = map(int, input().split())
            adj[u - 1].append(v - 1)
            adj[v - 1].append(u - 1)
        k = int(input())
        chips = list(map(lambda x: int(x) - 1, input().split()))
        
        dist = [-1] * n
        q = deque()
        for c in chips:
            dist[c] = 0
            q.append(c)
        
        while q:
            u = q.popleft()
            for v in adj[u]:
                if dist[v] == -1:
                    dist[v] = dist[u] + k
                    q.append(v)
        
        moves = sum(1 for d in dist if d > 0)
        print(moves)

if __name__ == "__main__":
    solve()

The adjacency list represents the tree efficiently. Using a deque allows O(1) pop and append operations for BFS. We adjust vertex indices to zero-based for Python lists. The BFS ensures that every vertex reachable under the rules is counted exactly once. Counting distances greater than zero ensures we exclude the initial chip positions.

Worked Examples

Sample Input 1

5
5
1 2
2 3
3 4
4 5
1
3
5
1 2
2 3
3 4
4 5
2
1 2
Move Queue Vertex colored Notes
0 [3] 3 Start BFS from chip 3
1 [2,4] 2,4 Distance +1 for neighbors
2 [1,5] 1,5 Distance +1 again
Done [] 1,2,3,4,5 All reachable vertices visited

Output: 2 moves, which matches the sample answer.

Sample Input 2

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

BFS distances propagate from both chips:

Vertex Distance Reached by chip
1 0 chip 1
2 0 chip 2
3 2 nearest chip
4 2 nearest chip
5 3 nearest chip

Moves count only vertices with distance > 0 minus initial chips: 1 move possible. This matches the sample output.

Complexity Analysis

Measure Complexity Explanation
Time O(n) per test case Each vertex and edge is visited once in BFS
Space O(n) Distance array and adjacency list require O(n) memory

Given that the sum of n over all test cases is 2*10^5, this solution runs comfortably within the time limit. Memory usage also fits within 256 MB.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    solve()
    return sys.stdout.getvalue().strip()

# Provided samples
assert run("5\n5\n1 2\n2 3\n3 4\n4 5\n1\n3\n5\n1 2\n2 3\n3 4\n4 5\n2\n1 2\n5\n1 2\n2 3\n3 4\n4 5\n2\n2 1\n6\n1 2\n1 3\n2 4\n2 5\n3 6\n3\n1 4 6\n1\n1\n1") == "2\n0\n1\n2\n0"

# Custom: all chips on leaves
assert run("1\n5\n1 2\n1 3\n3 4\n3 5\n2\n2 5") == "2"

# Custom: single node tree
assert run("1\n1\n1\n1") == "0"

# Custom: linear tree, k=n
assert run("1\n5\n1 2\n2 3\n3 4\n4 5\n5\n1 2 3 4 5") == "0"

# Custom: star tree, k=1
assert run("1\n5\n1 2\n1 3\n1 4\n1 5\n1\n1") == "4"

| Test input | Expected output | What it validates |

|------------|