[树的深度] Party

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 p_i (1 ≤ p_i ≤ n or p_i = -1). Every p_i denotes the immediate manager for the i-th employee. If p_i 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 (p_i ≠ 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.

Examples

Input

Output

Note

For the first example, three groups are sufficient, for example:

Employee 1
Employees 2 and 4
Employees 3 and 5

题意:给出n个点和他们的父结点,现在要将他们分成一些小组,小组内不能出现任何一个人的祖先,问最少可以分成几个组
思路:在一个组内不能出现一个点的祖先,也不能出现一个点的后代,因为如果出现了一个点的后代,则这个结点就是那些后代的祖先,这是不合法的
所以我们可以把深度相同的结点分位一组,最大深度就是要分的组数,因为题目的上下级关系可能会分成很多树形成一个森林,所以我们把每个结点当作树根来统计深度,最后保存一个最大的深度作为答案输出

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<queue>
 5 using namespace std;
 6 typedef long long ll;
 7 const int amn=1e5+5;
 8 int n,ans=0,m[amn],deep[amn],cnt;
 9 vector<int> eg[amn];
10 queue<int> q;
11 void bfs(int rt){
12     while(q.size())q.pop();q.push(rt);
13     memset(deep,0,sizeof deep);
14     deep[rt]=1;
15     cnt=0;
16     while(q.size()){
17         int u=q.front();q.pop();
18         cnt=max(cnt,deep[u]);
19         for(int i=0;i<eg[u].size();i++){
20             int v=eg[u][i];
21             deep[v]=deep[u]+1;
22             q.push(v);
23         }
24     }
25     ans=max(ans,cnt);
26 }
27 int main(){
28     scanf("%d",&n);
29     for(int i=1;i<=n;i++){
30         scanf("%d",&m[i]);
31         if(m[i]!=-1){
32             eg[m[i]].push_back(i);
33         }
34     }
35     ans=0;
36     for(int i=1;i<=n;i++){
37         bfs(i);
38     }
39     printf("%d
",ans);
40 }
41 /**
42 题意:给出n个点和他们的父结点,现在要将他们分成一些小组,小组内不能出现任何一个人的祖先,问最少可以分成几个组
43 思路:在一个组内不能出现一个点的祖先,也不能出现一个点的后代,因为如果出现了一个点的后代,则这个结点就是那些后代的祖先,这是不合法的
44 所以我们可以把深度相同的结点分位一组,最大深度就是要分的组数,因为题目的上下级关系可能会分成很多树形成一个森林,所以我们把每个结点当作树根来统计深度,最后保存一个最大的深度作为答案输出
45 **/