Codeforces Beta Round #87 (Div. 1 Only)A. Party（深搜）

Posted 2020-10-01

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Description

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 p_i (1 ≤ p_i≤ n or p_i = -1). Every pi denotes the immediate manager for the i-th employee. If p_i 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 (p_i ≠ 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.

Sample Input

5
-1
1
2
1
-1

Sample Output

3

思路

题解:

根据题意只要求出最大的树高即可。

 

#include<bits/stdc++.h>
using namespace std;
const int maxn =  2005;
vector<int>itv[maxn];
int res = 0;

void dfs(int u,int depth)
{
	res = max(res,depth);
	for (unsigned i = 0;i < itv[u].size();i++)
	{
		dfs(itv[u][i],depth + 1);
	}
}

int main()
{
	int n,p;
	scanf("%d",&n);
	for (int i = 1;i <= n;i++)
	{
		scanf("%d",&p);
		if (p == -1)	itv[0].push_back(i);
		else	itv[p].push_back(i);
	}
	dfs(0,0);
	printf("%d\n",res);
	return 0;
}

　

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
vector<int>itv[maxn];
int level[maxn];

int main()
{
	int n,p;
	scanf("%d",&n);
	for (int i = 1;i <= n;i++)
	{
		scanf("%d",&p);
		if (~p)	itv[p].push_back(i);
		else	itv[0].push_back(i);
	}
	queue<int>que;
	for (unsigned i = 0;i < itv[0].size();i++)
	{
		int u = itv[0][i];
		que.push(u);
		level[u] = 1;
		while (!que.empty())
		{
			u = que.front();
			que.pop();
			for (unsigned j = 0;j < itv[u].size();j++)
			{
				int v = itv[u][j];
				level[v] = level[u] + 1;
				que.push(v);
			}
		}
	}
	int res = 1;
	for (int i = 1;i <= n;i++)	res = max(res,level[i]);
	printf("%d\n",res);
	return 0;
}