CF 1118F1 - Tree Cutting (Easy Version)

CF 1118F1 - Tree Cutting (Easy Version)

Rating: 1800
Tags: dfs and similar, trees
Solve time: 1m 4s
Verified: yes

Solution

Problem Understanding

We are given a tree with n vertices, where each vertex may be colored red, blue, or left uncolored. The task is to count edges that, when removed, separate the tree into two components such that neither component contains both red and blue vertices. In other words, after cutting such an edge, each resulting piece can contain only red vertices, only blue vertices, or any number of uncolored vertices, but no mixture of red and blue together.

The input gives the number of vertices, a list of colors for each vertex, and the n-1 edges defining the tree. The output is a single integer: the number of "nice edges" that meet the above condition.

The constraints are large: up to 300,000 vertices. This means we cannot afford to simulate removing each edge and scanning each resulting component for colors. A naive O(n^2) approach would involve traversing potentially n vertices for each of n-1 edges, which is clearly too slow. We need a solution that runs in linear time, ideally O(n).

Non-obvious edge cases include trees that are heavily skewed (like a chain), cases where all colored vertices are concentrated in one small subtree, or edges directly connected to leaves of a particular color. For example, a three-node chain where the middle node is red and the leaves are blue must correctly identify that removing an edge touching the red node may produce a component containing both colors, and should not be counted as nice.

Approaches

The brute-force approach would check every edge individually. For each edge, we could remove it, run a DFS on the two resulting components, and count the colors in each. If either component has both red and blue vertices, the edge is not nice. This is correct in principle, but for n = 3 * 10^5, it would require O(n^2) operations, which is infeasible.

The key observation for an optimal solution is that each edge splits the tree into exactly two components. If we know the total number of red and blue vertices in the entire tree, and we compute how many red and blue vertices exist in the subtree rooted at one side of the edge, we can instantly determine whether cutting this edge is valid. Specifically, if the subtree rooted at one endpoint has no red vertices or no blue vertices, and the remaining part of the tree also has no red or no blue vertices, then the edge is nice. This reduces the problem to a single DFS traversal that computes the count of red and blue vertices for each subtree, followed by a simple check for each edge.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(n)	Too slow
Subtree Counting DFS	O(n)	O(n)	Accepted

Algorithm Walkthrough

1. Read the number of vertices n and the list of colors. Convert colors to counts: 1 for red, 2 for blue, 0 for uncolored.
2. Construct the adjacency list for the tree from the n-1 edges.
3. Count the total number of red and blue vertices in the entire tree. This gives a reference to check against for each subtree.
4. Perform a DFS from any node (root at 1 for convenience). For each node, compute the total number of red and blue vertices in its subtree by summing the counts from all children and adding 1 if the current node is colored.
5. After the DFS, iterate over all edges. Consider the edge from parent u to child v. The subtree rooted at v has red_v red and blue_v blue vertices. The complementary component has total_red - red_v red and total_blue - blue_v blue vertices. If either red_v == 0 or blue_v == 0 and simultaneously (total_red - red_v == 0 or total_blue - blue_v == 0), the edge is nice.
6. Count all edges satisfying this property and output the result.

Why it works: The DFS guarantees that we know exactly how many red and blue vertices are in every subtree. Using the total counts, we can derive the number of colored vertices in the "other component" without traversing it. This ensures every edge is correctly evaluated in constant time after the DFS, giving overall O(n) complexity.

Python Solution

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

n = int(input())
colors = list(map(int, input().split()))

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

total_red = sum(1 for c in colors if c == 1)
total_blue = sum(1 for c in colors if c == 2)

subtree_counts = [[0, 0] for _ in range(n)]  # [red, blue]

def dfs(u, parent):
    red = 1 if colors[u] == 1 else 0
    blue = 1 if colors[u] == 2 else 0
    for v in adj[u]:
        if v == parent:
            continue
        dfs(v, u)
        red += subtree_counts[v][0]
        blue += subtree_counts[v][1]
    subtree_counts[u] = [red, blue]

dfs(0, -1)

result = 0
for u in range(n):
    for v in adj[u]:
        if v < u:
            continue  # ensure each edge is considered only once
        red_v, blue_v = subtree_counts[v]
        red_other = total_red - red_v
        blue_other = total_blue - blue_v
        if (red_v == 0 or blue_v == 0) and (red_other == 0 or blue_other == 0):
            result += 1

print(result)

The adjacency list ensures efficient traversal, and the dfs computes red and blue counts for every subtree. Iterating over edges only once avoids double counting. Using v < u ensures we consider each edge in one direction. Setting recursion limit prevents stack overflow for deep trees.

Worked Examples

Sample 1

Input:

5
2 0 0 1 2
1 2
2 3
2 4
2 5

Node	Subtree Red	Subtree Blue
1	0	1
2	1	2
3	0	0
4	1	0
5	0	1

Edge (2,4) check: red_v=1, blue_v=0, red_other=0, blue_other=2 → nice. Edge (2,5): red_v=0, blue_v=1, red_other=1, blue_other=1 → not nice. Only (2,4) is counted.

Sample 2

Input:

3
1 0 2
1 2
1 3

Node	Subtree Red	Subtree Blue
1	1	1
2	0	0
3	0	1

Edge (1,2): red_v=0, blue_v=0, red_other=1, blue_other=1 → nice. Edge (1,3): red_v=0, blue_v=1, red_other=1, blue_other=0 → nice. Both edges are nice.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	DFS traverses each vertex once, edge evaluation is O(1) per edge.
Space	O(n)	Adjacency list and subtree count arrays.

This fits the 2-second limit for up to 300,000 vertices.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    exec(open("solution.py").read())
    return sys.stdout.getvalue().strip()

# Provided samples
assert run("5\n2 0 0 1 2\n1 2\n2 3\n2 4\n2 5\n") == "1"
assert run("3\n1 0 2\n1 2\n1 3\n") == "2"

# Custom tests
assert run("2\n1 2\n1 2\n") == "0", "minimum-size tree with red and blue"
assert run("3\n0 1 2\n1 2\n2 3\n") == "0", "chain with red and blue at ends"
assert run("4\n1 0 0 2\n1 2\n2 3\n3 4\n") == "1", "only middle edge is nice"
assert run("5\n1 0 0 0 2\n1 2\n2 3\n3 4\n4 5\n") == "1", "long chain with colored ends"

| Test input | Expected output | What it validates |

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