A. Party
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.
Examples
inputCopy
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
思路：题意是将所有雇员分成不同的组，保证每一组里的成员中不包括成员的领导。那么最少的分组方法就是同一级的分成一个组。那么就只用求层数最多的树的层数即可。
#include<iostream>
#include<cmath>
using namespace std;
int fath[2005];
int main()
{
int n,temp;
int ans = 0;
cin>>n;
for(int i = 1;i<=n;i++)
{
cin>>fath[i];
}
for(int i = 1;i<=n;i++)
{
temp = 0;
for(int j = i;j<=n&&j!=-1;j=fath[j])//对于每一个i，依次往上面查根节点，直到j==-1 求的该数的层数。
{
temp++;
}
ans = max(ans,temp);//不断更新最大的层数
}
cout<<ans<<endl;
return 0;
}