Algorithm

A. Party
time limit per test
3 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/115/A
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.

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



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>

int main(){

    int n(0); scanf("%d", &n);
    int *sup = new int[n + 1];

    for(int k = 1; k <= n; k++){scanf("%d", sup + k);}

    int total = 0;
    for(int k = 1; k <= n; k++){
        int temp(1), boss(sup[k]);
        while(boss != -1){boss = sup[boss]; ++temp;}
        if(temp > total){total = temp;}
    }

    printf("%d\n", total);

    delete[] sup;
    return 0;
}
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Input

x
–
+
cmd
5
-1
1
2
1
-1

Output

x
–
+
cmd
3
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Demonstration

Codeforces Solution-A. Party-Solution in C, C++, Java, Python

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