Tree Infection (Codeforces 1665C)
Tags: Greedy, trees

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. The parent of a vertex $$$v$$$ (different from root) is the previous to $$$v$$$ vertex on the shortest path from the root to the vertex $$$v$$$. Children of the vertex $$$v$$$ are all vertices for which $$$v$$$ is the parent.You are given a rooted tree with $$$n$$$ vertices. The vertex $$$1$$$ is the root. Initially, all vertices are healthy.Each second you do two operations, the spreading operation and, after that, the injection operation:   Spreading: for each vertex $$$v$$$, if at least one child of $$$v$$$ is infected, you can spread the disease by infecting at most one other child of $$$v$$$ of your choice.  Injection: you can choose any healthy vertex and infect it. This process repeats each second until the whole tree is infected. You need to find the minimal number of seconds needed to infect the whole tree.

Input: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. Description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the number of the vertices in the given tree.The second line of each test case contains $$$n - 1$$$ integers $$$p_2, p_3, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$), where $$$p_i$$$ is the ancestor of the $$$i$$$-th vertex in the tree.It is guaranteed that the given graph is a tree.It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case you should output a single integer — the minimal number of seconds needed to infect the whole tree.

Note: The image depicts the tree from the first test case during each second.A vertex is black if it is not infected. A vertex is blue if it is infected by injection during the previous second. A vertex is green if it is infected by spreading during the previous second. A vertex is red if it is infected earlier than the previous second.  Note that you are able to choose which vertices are infected by spreading and by injections.

Sample Inputs:
5
7
1 1 1 2 2 4
5
5 5 1 4
2
1
3
3 1
6
1 1 1 1 1

Sample Outputs:
4
4
2
3
4

Solution (Python):
from collections import Counter
import math
import heapq
n = int(input())
ans = []
for _ in range(n):
    m = int(input())
    temp = list(map(int, input().split()))
    storage = [1] + [0] * m
    for (idx, value) in enumerate(temp):
        storage[value] += 1
    storage = [x for x in storage if x]
    storage.sort(reverse=True)
    n = len(storage)
    storage = [-(x + i - n) for (i, x) in enumerate(storage)]
    heapq.heapify(storage)
    time = 0
    while storage[0] < time:
        mx = heapq.heappop(storage)
        heapq.heappush(storage, mx + 1)
        time -= 1
    ans.append(str(n - time))
print(' '.join(ans))

Solution (Python):
from collections import defaultdict
counter = defaultdict(int)

def solve(a):
    for ai in a:
        counter[ai] += 1
    count = list(counter.values())
    num_level = len(count)
    count.sort()
    for i in range(num_level):
        count[i] = max(count[i] - i - 2, 0)
    L = 0
    R = max(count)
    if R == 0:
        return num_level + 1

    def check(k):
        b = count.copy()
        for i in range(len(b)):
            b[i] = max(b[i] - k, 0)
        if sum(b) <= k:
            return True
        return False
    while R - L > 1:
        mid = (R + L) // 2
        if check(mid):
            R = mid
        else:
            L = mid
    return num_level + 1 + R
for a in [*open(0)][2::2]:
    counter.clear()
    res = solve(a.split())
    print(res)

Solution (Python):
import heapq
from re import T
tc = int(input())
for _ in range(tc):
    n = int(input())
    a = list(map(int, input().split()))
    tree = [0 for _ in range(n + 1)]
    tree[0] = 1
    for i in range(n - 1):
        tree[a[i]] += 1
    tree = [x for x in tree if x]
    n = len(tree)
    tree.sort(reverse=True)
    tree = [-(tree[i] - n + i) for i in range(n)]
    heapq.heapify(tree)
    time = 0
    while tree[0] < time:
        mx = heapq.heappop(tree)
        heapq.heappush(tree, mx + 1)
        time -= 1
    print(-time + n)