一次dfs就可以求解：

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 100010,M = 2 * N;
int h[N],e[M],ne[M],idx;
int n,m,res;

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

int dfs(int u){
    int a = 0,b = 0;//a、b分别表示u能到的叶节点的最远长度和次远长度
    for(int i = h[u];i != -1;i = ne[i]){
        int j = e[i];
        int d = dfs(j);
        if(d >= a) b = a,a = d;
        else if (d > b) b = d;
    }
    res = max(res,a + b);
    return a + 1; // j到u还有一条边
}

int main(){
    memset(h,-1,sizeof h);
    cin >> n >> m;
    for(int i = 2;i <= n + m;i++){
        int p;
        cin >> p;
        add(p,i);
    }
    dfs(1);
    cout << res << endl;

    return 0;
}

两次dfs求解：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;
const int N = 20010,M = 2 * N;
int h[N],e[M],ne[M],idx;
int d[N];
int n,m;

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

int bfs(int u){
    memset(d,-1,sizeof d);
    d[u] = 0;
    queue<int> q;
    q.push(u);

    while(q.size()){
        int t = q.front();
        q.pop();
        for(int i = h[t];i != -1;i = ne[i]){
            int j = e[i];

            if(d[j] == -1){
                d[j] = d[t] + 1;
                q.push(j);
            }
        }
    }
    int res = 1;
    for(int i = 2;i <= n + m;i++){
        if(d[res] < d[i]){
            res = i;
        }
    }
    return res;
}

int main(){
    memset(h,-1,sizeof h);
    cin >> n >> m;
    for(int i = 2;i <= n + m;i++){
        int p;
        cin >> p;
        add(p,i),add(i,p);
    }
    int l = bfs(1);
    int r = bfs(l);
    cout << d[r] << endl;

    return 0;
}