codeforces 115A 求最大树高 搜索
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题目:
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?
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.
Print a single integer denoting the minimum number of groups that will be formed in the party.
5-1121-1
3
For the first example, three groups are sufficient, for example:
- Employee 1
- Employees 2 and 4
- Employees 3 and 5
有一棵树,给出每个结点的父结点,问这棵树的最大层数是多少(保证输入是一棵树,即不存在环)
(即给一片森林,求所有树的最大高度
分析:
没有合并操作,不需要用并查集(也不知道怎样用
可以从叶节点往上搜,每上一层层数加一,取所有叶节点搜出来的层数的最大值就可以了
代码:
非递归:
#include<bits/stdc++.h>#include<queue>using namespace std;const int maxn=2e3+30;const __int64 inf=(__int64)1<<62;int fa[maxn];int n;int main() {while(cin>>n) {for(int i=1;i<=n;++i) cin>>fa[i];int ans=0;for(int i=1;i<=n;++i) {int j=i;int tmp=0;while(j!=-1) {tmp++;j=fa[j];}ans=max(ans,tmp);}cout<<ans<<endl;}return 0;}
参自dalao :http://blog.csdn.net/loy_184548/article/details/51442490
递归:
#include<bits/stdc++.h>#include<queue>using namespace std;const int maxn=2e3+30;const __int64 inf=(__int64)1<<62;int fa[maxn];int n;int ans,tmp;void dfs(int i) {if(i==-1) return;else {tmp++;dfs(fa[i]);}}int main() {while(cin>>n) {for(int i=1;i<=n;++i) cin>>fa[i];ans=0;for(int i=1;i<=n;++i) {tmp=0;dfs(i);ans=max(ans,tmp);}cout<<ans<<endl;}return 0;}
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