Codeforces 116C. Party 树的深度

C. Party

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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

树的深度即为答案。

#include 
#include 

using namespace std;

vectora[2222];
int p[2222]={0};
int n;
int deep[2222]={0};


void dfs(int i,int d)
{
    int len=a[i].size();
    deep[i]=d;
    for (int v=0;v>n;
    for (int i=1;i<=n;i++)
    {
        cin>>p[i];
        if (p[i]!=-1)
        {
            a[p[i]].push_back(i);
        }
    }
    for (int i=1;i<=n;i++)
    {
        if (p[i]==-1)
        {
            dfs(i,1);
        }
    }
    int ans=0;
    for (int i=1;i<=n;i++)
    {
        if (deep[i]>ans)
        {
            ans=deep[i];
        }
    }
    cout<