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.
Examples
input
Copy
5 -1 1 2 1 -1
output
Copy
3
Note
For the first example, three groups are sufficient, for example:
Employee 1
Employees 2 and 4
Employees 3 and 5
刚开始是没明白题，直接把每个节点拿来暴搜，妥妥TLE。把-1作为头开始搜更有效
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#define ll long long
usingnamespace std;
int n,depth=0;
vector<int> p,head;
map<int, vector<int>> mp;
void dfs(int i,int depth,int &res)
{
depth++;
res = max(depth, res);
if (mp[i].size() == 0)
{
return;
}
for (int k = 0; k < mp[i].size(); k++)
{
dfs(mp[i][k], depth, res);
}
}
int main()
{
cin >> n;
p.resize(n+1);
for (int i = 1; i < n+1; i++)
{
cin >> p[i];
if (mp.find(i) == mp.end())
{
mp[i] = vector<int>();
}
if (p[i] != -1)
{
if (mp.find(p[i]) == mp.end())
{
mp[p[i]] = vector<int>();
mp[p[i]].push_back(i);
}
else
{
mp[p[i]].push_back(i);
}
}
else
{
head.push_back(i);
}
}
int res = 0;
for (int i = 0; i <head.size() ; i++)
{
dfs(head[i], 0, res);
}
cout << res;
//system("pause");return0;
}
