codeforces --- 115A

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

思路：求树的高度。用并查集，不要压缩路径。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int father[2005], cnt;
int max(int x, int y)
{
    return x > y ? x : y;
}
void init(int n)
{
    for(int i = 1;i <= n;i ++)
        father[i] = i;
}

void find(int x)
{
    if(x == father[x])
        return;
    cnt ++;
    find(father[x]);
}

void unit(int x, int y)
{
    father[x] = y;
    return ;
}

int main(int argc, char const *argv[]) 
{
    int n, ans, temp;
    while(~scanf("%d", &n))
    {
        init(n);
        for(int i = 1;i <= n;i ++)
        {
            scanf("%d", &temp);
            if(temp != -1)
                unit(i, temp);
        }
        ans = 0;
        for(int i = 1;i <= n;i ++)
        {
            cnt = 0;
            find(i);
            ans = max(ans, cnt);
        }
        printf("%d\n", ans+1);
    }
    return 0;
}