codeforces-115A-Party
2016-05-18 11:18
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115A-Party
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
题目链接:cf-115A
题目大意:每个人有0或1个直系上司。开晚会分组,问最少问几组,使得每组里面成员没有上司关系(直接、间接)。
题目思路:刚开始以为是并查集,后来发现不是,其实就是几棵树,求最高的树的层数。简化一下,输入的是每个成员的上司即父节点编号。从叶子节点开始向上搜,找到最大的层数即是答案。
以下是代码:
// // 115A-Party.cpp // codeforce // // Created by pro on 16/5/18. // Copyright (c) 2016年 loy. All rights reserved. // #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #include<iomanip> using namespace std; int fa[2005]; int main() { int n; cin >> n; for(int i = 1; i <= n; i++) cin >> fa[i]; int ans= 0; for (int i = 1; i <= n; i++) { int tmp = 0; for (int j = i; j <= n && j != -1; j = fa[j]) { tmp++; } ans = max(ans,tmp); } cout << ans << endl; return 0; }
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