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Programming101
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HackerEarth The family tree of Bob problem solution

YASH PAL, 31 July 2024
In this HackerEarth The family tree of Bob problem solution Bob wants to know about his ancestors, therefore, he draws a graph of his family. In that graph, root is the eldest-known family member. The leaf node is a member who has no children.
You are given a Q query and N family members. They have to just print the kth ancestors with respect to that query. A member can have only one parent.
Note: Print -1 if the query is incorrect, that is, if the kth ancestor is not available. The root of the tree is 1.
HackerEarth The family tree of Bob problem solution

HackerEarth The family tree of Bob problem solution.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 5e5 + 14, lg = 19;
int n, h[maxn], par[lg][maxn];
vector<int> g[maxn];
void dfs(int v = 0, int p = -1){
for(int u : g[v])
if(u != p){
par[0][u] = v;
h[u] = h[v] + 1;
dfs(u, v);
}
}

int parat(int v, int h){
for(int i = 0; i < lg; i++)
if(h >> i & 1)
v = par[i][v];
return v;
}

int main(){
ios::sync_with_stdio(0), cin.tie(0);
int q;
cin >> n >> q;
for(int i = 0; i < n - 1; i++){
int v, u;
cin >> v >> u;
v--, u--;
g[v].push_back(u);
g[u].push_back(v);
}
dfs();
for(int k = 1; k < lg; k++)
for(int i = 0; i < n; i++)
par[k][i] = par[k - 1][ par[k - 1][i] ];
while(q--){
int v, k;
cin >> v >> k;
v--;
cout << (h[v] < k ? -1 : parat(v, k) + 1) << 'n';
}
}

Second solution

import java.io.*;
import java.util.*;

public class Main implements Runnable
{
int depth = 0;
public static void main(String args[]) throws Exception
{
new Thread(null, new Main(), "Main", 1<<26).start();
}
public void run()
{
try
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
String s1[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(s1[0]);
int m = Integer.parseInt(s1[1]);
ArrayList<Integer>[] g = new ArrayList[n];
for (int i = 0; i < n; i++)
g[i] =new ArrayList<Integer>();

for(int i = 0; i < n - 1; i++)
{
String s2[] = br.readLine().trim().split(" ");
int u = Integer.parseInt(s2[0])-1;
int v = Integer.parseInt(s2[1])-1;
g[u].add(v);
g[v].add(u);
}

ArrayList<Query>[] qu = new ArrayList[n];
for (int i = 0; i < n; i++)
qu[i] =new ArrayList<Query>();

for(int i = 0; i < m; i++)
{
String s2[] = br.readLine().trim().split(" ");
int x = Integer.parseInt(s2[0])-1;
int y = Integer.parseInt(s2[1]);
qu[x].add(new Query(y, i));
}

int ans[] = new int[n+5];
int sol[] = new int[m];
depth = 0;
dfs(g, qu, 0, -1, ans, sol);

for(int i = 0; i < m; i++)
bw.write(sol[i]+1 + "n");
bw.flush();
}
catch(Exception e)
{
System.out.println("Error!"+"n");
e.printStackTrace();
System.exit(0);
}

}

public void dfs(ArrayList<Integer>[] g, ArrayList<Query> qu[], int s, int par, int ans[], int sol[])
{
ans[depth++] = s;
for(int i = 0; i < qu[s].size(); i++)
if((depth - qu[s].get(i).x - 1) < 0)
sol[qu[s].get(i).id] = -2;
else
sol[qu[s].get(i).id] = ans[depth - qu[s].get(i).x - 1];

for(int i = 0; i < g[s].size(); i++)
if(g[s].get(i) != par)
dfs(g, qu, g[s].get(i), s, ans, sol);

depth--;

}

}

class Query
{
int x;
int id;
Query(int x, int id)
{
this.x = x;
this.id = id;
}
}
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