# HackerEarth Oliver and the Game problem solution

In this HackerEarth Oliver and the Game problem solution Oliver and Bob are best friends. They have spent their entire childhood in the beautiful city of Byteland. The people of Byteland live happily along with the King.
The city has a unique architecture with total N houses. The King's Mansion is a very big and beautiful bungalow having address = 1. Rest of the houses in Byteland have some unique address, (say A), are connected by roads and there is always a unique path between any two houses in the city. Note that the King's Mansion is also included in these houses.

Oliver and Bob have decided to play Hide and Seek taking the entire city as their arena. In the given scenario of the game, it's Oliver's turn to hide and Bob is supposed to find him.
Oliver can hide in any of the houses in the city including the King's Mansion. As Bob is a very lazy person, for finding Oliver, he either goes towards the King's Mansion (he stops when he reaches there), or he moves away from the Mansion in any possible path till the last house on that path.

Oliver runs and hides in some house (say X) and Bob is starting the game from his house (say Y). If Bob reaches house X, then he surely finds Oliver.

Given Q queries, you need to tell Bob if it is possible for him to find Oliver or not.

The queries can be of the following two types:
0 X Y : Bob moves towards the King's Mansion.
1 X Y : Bob moves away from the King's Mansion

## HackerEarth Oliver and the Game problem solution.

`#include<bits/stdc++.h>using namespace std;int vertex;vector<vector<int> > tree; //used for representing the treevector<bool> visited;vector<int> starttime; // starttime[i] notes the time at which DFS enters node ivector<int> endtime; // endtime[i] notes the time at which DFS exits node iint timer = 0; // a global variable that stores the timer at that instantvoid  makeTree()  // takes the input and creates a directed graph representing the tree{    scanf("%d",&vertex);    tree.resize(vertex+1);    for(int i = 1; i < vertex ; i++)    {   int x,y;        scanf("%d%d",&x,&y);        tree[x].push_back(y);    }}void measureTime(int v) // Performs Depth First Search{    visited[v] = 1;    starttime[v] = timer++;    for(int i = 0 ; i < tree[v].size() ; i++) // calling measureTime() for adjacent nodes of node v and performing DFS    {        if( visited [ tree[ v ][ i ] ] == 0  )            measureTime(tree[v][i]);    }    endtime[v] = timer++;}int check(int x, int y){    if( starttime[x] > starttime[y] && endtime[x] < endtime[y] ) // checks weather node x lies in the subtree of node y or not        return 1;    return 0;}int main(){    makeTree();    visited.resize(vertex+1,0);    starttime.resize(vertex+1,0);    endtime.resize(vertex+1,0);    measureTime(1);    int q;    scanf("%d",&q);    while(q--)    {        int type,x,y;    scanf("%d%d%d",&type,&x,&y);        if( !check(x,y) && !check(y,x) )        {            printf("NO\n");            continue;        }        if(type == 0)        {            if(check(y,x) == 1)                printf("YES\n");            else                printf("NO\n");        }        else        {            if(check(x,y) == 1)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}`

### Second solution

`#include <algorithm>#include <bitset>#include <cassert>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <deque>#include <functional>#include <iostream>#include <iomanip>#include <list>#include <math.h>#include <map>#include <numeric>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <utility>#include <vector>#define LL long long#define ULL unsigned long long#define F first#define S second#define pb push_back#define FOR(i,lb,ub) for(i=lb;i<=ub;i++)#define RFOR(i,ub,lb) for(i=ub;i>=lb;i--)#define FORS(it,v) for(it=v.begin();it!=v.end();it++)using namespace std;vector<int> G[100005];bool visited[100005];int in[100005], out[100005], ts;int cnt;void dfs(int x){  cnt++;  visited[x] = true;  in[x] = ++ts;  for (int i = 0; i <G[x].size(); ++i)  {    if (!visited[G[x][i]])      dfs(G[x][i]);  }  out[x] = ++ts;}bool is_subtree(int x, int y)//if y lies in subtree of x{  if (in[x] <= in[y] && out[x]>=out[y])    return true;  return false;}int main(){    ios_base::sync_with_stdio(false);    cin.tie(NULL);    int t,i,j;    int n;    cin>>n;    assert(n>=1 && n<=100000);    FOR(i,0,n-2)    {      int a,b;      cin>>a>>b;      assert(a>=1 && a<=n);      assert(b>=1 && b<=n);      G[a].pb(b);      G[b].pb(a);    }    dfs(1);    assert(cnt==n);    int q;    cin>>q;    assert(q>=1 && q<=500000);    while (q--)    {      int a,x,y;      cin>>a>>x>>y;      assert(a==0 || a==1);      assert(x>=1 && x<=n);      assert(y>=1 && y<=n);      if ((!a && is_subtree(x,y)) || (a && is_subtree(y,x)))        cout<<"YES\n";      else        cout<<"NO\n";     }        return 0;}`