Codeforces 115A- Party

A. 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 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.

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

题意：求数的最大深度

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>

using namespace std;

typedef long long LL;
const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);

int n,a[2010],ans,tem;
void dfs(int u)
{
tem++;
if(a[u]==-1)
return ;
dfs(a[u]);
}
void solve()
{
ans=-INF;
for(int i=1;i<=n;i++)
{
tem=0;
dfs(i);
ans=max(ans,tem);
}
printf("%d\n",ans);
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
scanf("%d",a+i);
solve();
}
return 0;
}