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

題目大意：每個人有0或1個直系上司。開晚會分組，問最少問幾組，使得每組裡面成員沒有上司關係（直接、間接）。

題目思路：剛開始以為是並查集，後來發現不是，其實就是幾棵樹，求最高的樹的層數。簡化一下，輸入的是每個成員的上司即父節點編號。從葉子節點開始向上搜，找到最大的層數即是答案。

以下是程式碼：

//
//  115A-Party.cpp
//  codeforce
//
//  Created by pro on 16/5/18.
//  Copyright (c) 2016年 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;
}