codeforces --- 115A
3 seconds
256 megabytes
standard input
standard output
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
-1
1
2
1
-1
3 思路:求树的高度。用并查集,不要压缩路径。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int father[], cnt;
int max(int x, int y)
{
return x > y ? x : y;
}
void init(int n)
{
for(int i = ;i <= n;i ++)
father[i] = i;
} void find(int x)
{
if(x == father[x])
return;
cnt ++;
find(father[x]);
} void unit(int x, int y)
{
father[x] = y;
return ;
} int main(int argc, char const *argv[])
{
int n, ans, temp;
while(~scanf("%d", &n))
{
init(n);
for(int i = ;i <= n;i ++)
{
scanf("%d", &temp);
if(temp != -)
unit(i, temp);
}
ans = ;
for(int i = ;i <= n;i ++)
{
cnt = ;
find(i);
ans = max(ans, cnt);
}
printf("%d\n", ans+);
}
return ;
}
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