codeforces-115A-Party

115A-Party
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
Employee A is the immediate manager of employee B Employee B has an immediate manager employee C such that employee A is the superior of employee C. The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output Print a single integer denoting the minimum number of groups that will be formed in the party.
input 5 -1 1 2 1 -1 output 3
제목 링크: cf - 115A
제목: 1 인당 0 또는 1 명의 직계 상사 가 있다.파 티 를 열 어 조 를 나 누 어 최소 몇 조 에 게 물 어 봐 서 각 조 의 구성원 들 이 상사 관계 가 없 게 한다 (직접, 간접).
제목 사고: 처음에 집 이 라 고 생각 했 는데 아니 라 는 것 을 알 게 되 었 습 니 다. 사실은 몇 그루 의 나무 이 고 가장 높 은 나무의 층 수 를 구 했 습 니 다.간단하게 말하자면, 각 구성원 의 상사 즉 부모 노드 번 호 를 입력 한다.잎 사 귀 노드 부터 위로 뒤 져 가장 큰 층 수 를 찾 는 것 이 답 이다.
다음은 코드:

//
// 115A-Party.cpp
// codeforce
//
// Created by pro on 16/5/18.
// Copyright (c) 2016  loy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>
using namespace std;
int fa[2005];
int main()
{
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> fa[i];
    int ans= 0;
    for (int i = 1; i <= n; i++)
    {
        int tmp = 0;
        for (int j = i; j <= n && j != -1; j = fa[j])
        {
            tmp++;
        }
        ans = max(ans,tmp);
    }
    cout << ans << endl;
    return 0;
}