[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

5
-1
1
2
1
-1

Output

3

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

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
typedef long long ll;
const int amn=1e5+5;
int n,ans=0,m[amn],deep[amn],cnt;
vector<int> eg[amn];
queue<int> q;
void bfs(int rt){
    while(q.size())q.pop();q.push(rt);
    memset(deep,0,sizeof deep);
    deep[rt]=1;
    cnt=0;
    while(q.size()){
        int u=q.front();q.pop();
        cnt=max(cnt,deep[u]);
        for(int i=0;i<eg[u].size();i++){
            int v=eg[u][i];
            deep[v]=deep[u]+1;
            q.push(v);
        }
    }
    ans=max(ans,cnt);
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&m[i]);
        if(m[i]!=-1){
            eg[m[i]].push_back(i);
        }
    }
    ans=0;
    for(int i=1;i<=n;i++){
        bfs(i);
    }
    printf("%d
",ans);
}
/**
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
**/