Problem: Given an undirected tree, let the distance between vertices u and v be the number of edges on the simple path from u to v. The diameter of a tree is the maximum among the distances between any two vertices. We will call a tree _good_ if and only if its diameter is at most K.  You are given an undirected tree with N vertices numbered 1 through N. For each i (1≦i≦N-1), there is an edge connecting vertices A_i and B_i.  You want to remove zero or more vertices from the tree, so that the resulting tree is good. When a vertex is removed, all incident edges will also be removed. The resulting graph must be connected.  Find the minimum number of vertices that you need to remove in order to produce a good tree.

Submission attempts (in order):
import sys
from collections import deque


def bfs(links, s, limit):
    not_reachable = (1 << len(links)) - 1
    q = deque([(0, s, -1)])
    pop = q.popleft
    extend = q.extend
    while q:
        cost, v, p = pop()
        if cost > limit:
            break
        not_reachable ^= 1 << v
        cost += 1
        extend((cost, u, v) for u in links[v] if u != p)
    return not_reachable


def solve(n, k, ab):
    if k == 1:
        return n - 2

    links = [set() for _ in range(n)]
    link_list = []
    for a, b in zip(ab[0::2], ab[1::2]):
        a -= 1
        b -= 1
        links[a].add(b)
        links[b].add(a)
        link_list.append((a, b))

    limit = k // 2
    if k % 2 == 0:
        ans = min(bin(bfs(links, v, limit)).count('1') for v in range(n))
    else:
        dists = [bfs(links, v, limit) for v in range(n)]
        ans = min(bin(dists[a] & dists[b]).count('1') for a, b in link_list)
    return ans


n, k = map(int, sys.stdin.buffer.readline().split())
ab = map(int, sys.stdin.buffer.read().split())
ab = list(zip(ab, ab))
print(solve(n, k, ab))


import sys
from collections import deque, defaultdict


def bfs(links, s, limit):
    not_reachable = [True] * len(links)
    q = deque([(0, s, -1)])
    while q:
        cost, v, p = q.popleft()
        if cost > limit:
            break
        not_reachable[v] = False
        cost += 1
        q.extend((cost, u, v) for u in links[v] if u != p)
    return not_reachable


def solve(n, k, ab):
    if k == 1:
        return n - 2

    links = [set() for _ in range(n)]
    for a, b in zip(ab[0::2], ab[1::2]):
        a -= 1
        b -= 1
        links[a].add(b)
        links[b].add(a)

    ans = n
    thr = k // 2
    if k % 2 == 0:
        ans = min(sum(bfs(links, v, thr) for v in range(n)))
    else:
        dists = [bfs(links, v, thr) for v in range(n)]
        for a, b in zip(ab[0::2], ab[1::2]):
            a -= 1
            b -= 1
            ans = min(ans, sum(d1 and d2 for d1, d2 in zip(dists[a], dists[b])))
    return ans


n, k, *ab = map(int, sys.stdin.buffer.read().split())
print(solve(n, k, ab))


import sys
from collections import deque


def bfs(links, s, limit, bits):
    not_reachable = (1 << len(links)) - 1
    q = deque([(0, s, -1)])
    while q:
        cost, v, p = q.popleft()
        if cost > limit:
            break
        not_reachable ^= bits[v]
        cost += 1
        for u in links[v]:
            if u != p:
                q.append((cost, u, v))
    return not_reachable


def solve(n, k, ab):
    if k == 1:
        return n - 2

    links = [set() for _ in range(n)]
    for a, b in ab:
        a -= 1
        b -= 1
        links[a].add(b)
        links[b].add(a)

    limit = k // 2
    bits = [1 << i for i in range(n)]
    if k % 2 == 0:
        ans = min(bin(bfs(links, v, limit, bits)).count('1') for v in range(n))
    else:
        dists = [bfs(links, v, limit, bits) for v in range(n)]
        ans = min(bin(dists[a - 1] & dists[b - 1]).count('1') for a, b in ab)
    return ans


n, k = map(int, sys.stdin.buffer.readline().split())
ab = map(int, sys.stdin.buffer.read().split())
ab = list(zip(ab, ab))
print(solve(n, k, ab))


Verdicts: Runtime Error

Runtime Error

Accepted