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