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

传送门

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。

 

+ View Code?

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#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; }

　

+ View Code?

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#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; }