Codeforces #115 A 树+最大层数

   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.

input
5
-1
1
2
1
-1

output
3

题目大意：

输入n个员工，编号从1到n。接着是n行，分别为这n个员工的直系经理的编号，如果为-1，则说明此员工没有直系经理。

用树的图形来解此题，求树的最高层数。因为我们输入的是每个成员的上司编号，即父节点的编号，所以只需从叶子节点开始向上搜，求出最大的层数即可。

代码实现：

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

int father[2010];

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; i++)
            scanf("%d", &father[i]);      //输入经理的编号
        int ans;
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            ans = 0;                    //记录树的深度
            for (int j = i; j <= n; j = father[j]) {   //j=father[j]用于判断是否已到根节点，如果到了根节
                                                                 //点，则循环结束，此时的ans就是深度。
                if (j != -1)
                    ans++;
                else
                    break;
            }
            cnt = max(cnt, ans);          //更新数据，找到最大的那个值即是答案
        }
        printf("%d\n", cnt);
    }
    return 0;
}