//#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e3 + 10;
ll n, ans[N], dp[N << 1], minn = 0x3f3f3f3f;
vector<pair<int, int> > vec[N << 1];
inline ll read(){
	ll x = 0, m = 1;
	char ch = getchar();
	while(!isdigit(ch)){
		if(ch == '-') m = -1;
		ch = getchar();
	}
	while(isdigit(ch)){
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * m;
}
inline void write(ll x){
	if(x < 0){
		putchar('-');
		write(-x);
		return;
	}
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
inline int dfs(int u, int fa, int fr){
	if(fr && dp[fr]) return dp[fr];
	priority_queue<int> q;
	for(int i = 0; i < vec[u].size(); ++ i){
		int v = vec[u][i].first, fr2 = vec[u][i].second;
		if(v == fa) continue;
		q.push(dfs(v, u, fr2));
	}
	int Ans = 0;
	for(int i = 1; !q.empty(); ++ i, q.pop()){
		Ans = max(Ans, q.top() + i);
	}
	return dp[fr] = Ans;
}
signed main(){
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
    n = read();
    int cnt = 0;
    for(int i = 2; i <= n; ++ i){
    	int x = read();
    	++ cnt, vec[x].push_back(make_pair(i, cnt));
    	++ cnt, vec[i].push_back(make_pair(x, cnt));
	}
	for(int i = 1; i <= n; ++ i){
		ans[i] = dfs(i, 0, 0);
		if(ans[i] < minn) minn = ans[i];
	}
	write(minn + 1), putchar('\n');
	for(int i = 1; i <= n; ++ i){
		if(minn == ans[i]){
			write(i), putchar(' ');
		}
	}
	return 0;
}

为什么我的 $minn$ 输出的时候要 $+1$ ?

$O(n \log n)$ 的做法。