Codeforces 116C. Party 树的深度

最新推荐文章于 2021-08-11 23:52:27 发布

cyendra

最新推荐文章于 2021-08-11 23:52:27 发布

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公司员工分为无上级和有上级两类，无上级的员工可以领导有上级的员工，但不允许形成管理环。现在要将所有员工按没有上下级关系的原则分组，求最少能分成多少组。输入包含员工数和每个员工的直接上级，输出最少的分组数。第一例中，至少需要3组：员工1一组，员工2和4一组，员工3和5一组。

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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 <iostream>
#include <vector>

using namespace std;

vector<int>a[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<len;v++)
    {
        dfs(a[i][v],d+1);
    }
}

int main()
{
    cin>>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<<ans<<endl;
    return 0;
}