Codeforces 115A- Party（DFS）

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

题意：给一片森林，求当中的树的最大深度。
。
爆搜就可以

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
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
",ans);
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
			scanf("%d",a+i);
		solve();
	}
	return 0;
}