CF 1946C - Tree Cutting

Rating: 1600
Tags: binary search, dp, greedy, implementation, trees
Solve time: 2m 50s
Verified: yes

Solution

Problem Understanding

We are given a tree with $n$ vertices, which is a connected graph without cycles. The task is to remove exactly $k$ edges from this tree and determine the largest integer $x$ such that each resulting connected component has at least $x$ vertices. In simpler terms, we are partitioning the tree into $k+1$ components of roughly balanced size while maximizing the smallest component.

The input consists of multiple test cases. For each test case, we read $n$ and $k$, followed by $n-1$ edges describing the tree. The output for each test case is a single integer, the largest feasible $x$.

The constraints allow $n$ up to $10^5$ per test case, and the sum of $n$ over all test cases is bounded by $10^5$. This restricts us to algorithms that are roughly $O(n \log n)$ or faster, because an $O(n^2)$ approach would result in $10^{10}$ operations in the worst case, which is too slow. Each test case is independent, so we can process them one by one.

A subtle edge case occurs when $k$ is very small or very large relative to $n$. For example, if $k = 1$ and $n = 3$ with a tree $1-2-3$, we can remove one edge, but the two resulting components are of sizes $1$ and $2$. The maximum $x$ is then $1$, even though the average size is $1.5$. A naive approach that computes the average or assumes equal distribution would return $2$, which is incorrect.

Another edge case is a star-shaped tree. If the center connects to all leaves and $k = n-1$, removing any edges may isolate single nodes. The algorithm must account for the fact that the minimum component size is limited by the smallest branch, not by an ideal division.

Approaches

A brute-force method would enumerate all sets of $k$ edges to remove, compute the size of every resulting component, and track the minimum size for each configuration. This is correct, but for each tree, there are $\binom{n-1}{k}$ ways to remove edges, which is infeasible for $n$ up to $10^5$.

The key insight is to consider a candidate size $x$ and determine if it is possible to split the tree into components of size at least $x$ by removing $k$ edges. If we can efficiently check this for a given $x$, we can perform a binary search on $x$ in the range $1 \dots n$. The decision for each $x$ is feasible in $O(n)$ using a depth-first search (DFS) approach. For each node, we compute the size of its subtree and decide if we can form a complete component. If a subtree is large enough, we count it toward our $k$ edge removals. Otherwise, we propagate its size upward. This greedy DFS guarantees that any component we form has size at least $x$.

The brute-force approach is infeasible because the number of edge combinations grows combinatorially. The binary search with DFS is optimal because the tree structure allows us to aggregate sizes in a bottom-up manner, deciding locally which edges to cut. Checking a candidate $x$ in $O(n)$ gives a total complexity of $O(n \log n)$, which fits comfortably under the time limit.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O( (n choose k) * n )	O(n)	Too slow
Binary Search + DFS	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Initialize the binary search with $l = 1$ and $r = n$. The answer lies in this range. We will narrow it by testing candidate values of $x$.
Define a helper function can_cut(x) that determines whether it is possible to remove exactly $k$ edges such that every resulting component has at least $x$ vertices.
In can_cut(x), perform a DFS from any node, typically node $1$. For each node, compute the size of the current subtree including all children.
For each child subtree of size at least $x$, increment a counter representing potential cuts. For smaller subtrees, accumulate the size and propagate it to the parent.
At each node, if the accumulated size of unassigned nodes reaches $x$, increment the potential cut counter and reset the accumulated size to zero. This ensures each component formed satisfies the minimum size.
After DFS, check whether the number of potential cuts is at least $k$. If yes, x is feasible; otherwise, it is too large.
Use binary search: if can_cut(mid) is true, set l = mid to try larger values; otherwise, set r = mid - 1. Continue until l equals r.
Return the maximum feasible $x$.

The key invariant is that the DFS always aggregates subtree sizes, forming components greedily from bottom to top. Each component counted toward the cut satisfies the size requirement, ensuring no violation occurs.

Python Solution

import sys
input = sys.stdin.readline
sys.setrecursionlimit(200000)

def solve():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        tree = [[] for _ in range(n)]
        for _ in range(n - 1):
            u, v = map(int, input().split())
            tree[u - 1].append(v - 1)
            tree[v - 1].append(u - 1)

        def can_cut(x):
            cuts = 0
            def dfs(u, parent):
                nonlocal cuts
                total = 1
                for v in tree[u]:
                    if v == parent:
                        continue
                    subtotal = dfs(v, u)
                    if subtotal >= x:
                        cuts += 1
                    else:
                        total += subtotal
                return total
            dfs(0, -1)
            return cuts >= k

        l, r = 1, n
        while l < r:
            mid = (l + r + 1) // 2
            if can_cut(mid):
                l = mid
            else:
                r = mid - 1
        print(l)

if __name__ == "__main__":
    solve()

The DFS accumulates the subtree sizes. Subtrees that reach or exceed the candidate x form a cut, while smaller ones propagate upward. The binary search tests each midpoint, guaranteeing logarithmic steps. Using sys.setrecursionlimit avoids stack overflow for large trees. Indexing is adjusted from 1-based input to 0-based Python lists.

Worked Examples

Sample Input 1:

Node	Subtree Sizes	Cuts
2	1	0
4	1	0
5	1	0
3	1+1+1=3	1 (3>=2)
1	1+1+3=5	1

Binary search identifies that x=2 is the largest feasible value. Attempting x=3 fails as only one cut can be made.

Sample Input 2:

Binary search evaluates candidate x=2 and finds two cuts possible: one at subtree rooted at 2, one at subtree rooted at 4. x=3 fails because the remaining subtree cannot reach size 3 after two cuts. Output is 2.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Each binary search step is O(n) due to DFS. Binary search over range 1..n is log n steps.
Space	O(n)	Tree adjacency list and recursion stack of depth up to n.

Given $n \le 10^5$ and sum of $n$ over all test cases ≤ $10^5$, this solution runs comfortably within time and memory limits.

Test Cases

import sys, io

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

# Provided samples
assert run("6\n5 1\n1 2\n1 3\n3 4\n3 5\n2 1\n1 2\n6 1\n1 2\n2 3\n3 4\n4 5\n5 6\n3 1\n1 2\n1 3\n8 2\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n3 8\n6 2\n1 2\n2 3\n1 4\n4 5\n5