蛮不错的一道题目，你要利用树的性质分析出，你只需要维护上一次的树的直径的两个端点就好了

#include<iostream>

using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 6e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}


int n,q,m;

// int e[N],ne[N],w[N],h[N],idx;
// void add(int a,int b,int c){
	// e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
// }

int dep[N];
int fa[N][22];


int lca(int a,int b){
	if(dep[a]<dep[b])swap(a,b);
	
	for(int i=20;i>=0;--i)if(dep[fa[a][i]]>=dep[b])a = fa[a][i];
	
	if(a==b)return a;
	for(int i=20;i>=0;--i)if(fa[a][i]!=fa[b][i])a = fa[a][i],b =fa[b][i];
	return fa[a][0];
	
}

int dist(int a,int b){
	return dep[a]+dep[b]-2*dep[lca(a,b)];
}

void solve()
{
	cin>>n;
	dep[2] = dep[3] = dep[4] = 2;
	dep[1] = 1;
	fa[1][0] = 0;
	fa[2][0] = fa[3][0] = fa[4][0] = 1;
	int tem = 4;
	int A = 2,B = 3;
	
	while(n--){
		int a;cin>>a;
		int b = ++tem,c = ++tem;
		fa[b][0] = a,fa[c][0] = a;
		dep[b] = dep[a]+1,dep[c] = dep[a]+1;
		for(int i=1;i<=20;i++)fa[b][i] = fa[fa[b][i-1]][i-1] ;
		for(int i=1;i<=20;i++)fa[c][i] = fa[fa[c][i-1]][i-1] ;
		int dista = dist(b,A),distb = dist(b,B);
		int dists = dist(A,B);
		//cout<<A<<" "<<B<<" "<<b<<" "<<dista<<" "<<distb<<" "<<dists<<"\n";
		if(dista<=dists&&distb<=dists){cout<<dists<<"\n";continue;}
		if(dista>dists&&dista>=distb){B=b;cout<<dista<<"\n";continue;}
		if(distb>dists){A=b;cout<<distb<<"\n";continue;}
		
		
		
	}

}

signed main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _;
	//cin>>_;
	_ = 1;
	while(_--)solve();
	return 0;
}