HackerEarth Tree queries problem solution

In this HackerEarth Tree queries problem solution You are given a tree T with N nodes rooted at node 1. Each node has a value A[i] associated with it.

You are required to perform Q queries where each query is of type:

1. u x: Increment the value associated with node u by x.
2. p: Find the shortest distance of any node from the root that has a value strictly greater than p. If no such node exists, then print -1.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <iomanip>
#include <fstream>
#include <stack>

using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll , ll> PLL;
typedef long double ld;
 
#define pb push_back
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
int mod = 1000000007;
const int inf = 1034567891;
const ll LL_INF = 1234567890123456789ll;
#define PI 3.14159265
#define endl 'n'
#define F first
#define S second
#define debug(x) cout << #x << " = " << x << endl;
#define TRACE
 
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
#define out(container) for (auto it : container) cout << it << " "; cout << endl;
 
 
template < typename T > T GCD(T a, T b)            { ll t; while(a) { t = a; a = b % a; b = t; } return b; }
template < typename T > string toString(T a)       { return to_string(a); }
template < typename T > void toInt(string s, T &x) { stringstream str(s); str >> x;}
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % mod;}
inline int powr(int a, ll b){
  int x = 1 % mod;
  while(b){
    if(b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a){ return powr(a, mod - 2);}

const int N = 2e5 + 5;

ll t[4*N], a[N];

void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = a[tl];
  } else {
    int tm = (tl + tr) / 2;
    build(v*2, tl, tm);
    build(v*2+1, tm+1, tr);
    t[v] = max(t[v*2], t[v*2+1]);
  }
}
void update(int v, int tl, int tr, int pos, ll new_val) {
  if (tl == tr) {
    t[v] += new_val;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      update(v*2, tl, tm, pos, new_val);
    else
      update(v*2+1, tm+1, tr, pos, new_val);
    t[v] = max(t[v*2], t[v*2+1]);
  }
}

int get_first(int v, int lv, int rv, int l, int r, ll x) {
  if(lv > r || rv < l) return -1;
  if(l <= lv && rv <= r) {
    if(t[v] <= x) return -1;
    while(lv != rv) {
      int mid = lv + (rv-lv)/2;
      if(t[2*v] > x) {
        v = 2*v;
        rv = mid;
      } else {
        v = 2*v+1;
        lv = mid+1;
      }
    }
    return lv;
  }
  int mid = lv + (rv-lv)/2;
  int rs = get_first(2*v, lv, mid, l, r, x);
  if(rs != -1) return rs;
  return get_first(2*v+1, mid+1, rv, l ,r, x);
}

vector<int> adj[N];
bool vis[N];
int dis[N];

void dfs(int s, int d) {
  vis[s] = true;
  dis[s] = d;
  for (auto it : adj[s]) {
    if (!vis[it]) {
      dfs(it, d + 1);
    }
  }
}

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt;
  cin >> tt;
  assert(1 <= tt and tt <= 10);
  while(tt--){
      for(int i = 0 ; i < N ; i++)
      {
        adj[i].clear();
        vis[i] = false;
        dis[i] = 0;
      }

      for(int i = 0 ; i < 4*N ; i++) t[i] = 0;

      int n;
      cin >> n;
      assert(2 <= n and n <= 100000);
      ll val[N];
      for (int i = 1; i <= n; i++) {
        cin >> val[i];
        assert(1 <= val[i] and val[i] <= 1000000);
      }  
      ll u, v;
      for (int i = 0; i < n - 1; i++) {
        cin >> u >> v;
        assert(1 <= u and u <= n);
        assert(1 <= v and v <= n);
        adj[u].pb(v);
        adj[v].pb(u);
      }
      dfs(1, 0);
      vector<PLL> vec;
      vector<int> pos(n + 1, 0);
      for (int i = 1; i <= n; i++) {
        vec.pb({dis[i], i});
      }
      sort(all(vec));
      int m = n;
      for (int i = 1; i <= m; i++) {
        pos[vec[i - 1].S] = i;
        a[i] = val[vec[i - 1].S];
      }
      build(1, 1, m);
      int q;
      cin >> q;
      assert(2 <= q and q <= 100000);
      while (q--) {
        int type;
        cin >> type;
        if (type == 1) {
          ll x, y;
          cin >> x >> y;
          assert(1 <= x and x <= n);
          assert(1 <= y and y <= 1000000);
          update(1, 1, m, pos[x], y);
          val[x] += y;
          a[pos[x]] += y;
        } else {
          ll x;
          cin >> x;
          assert(1 <= x and x <= 1000000000000);
          int ind = get_first(1, 1, m, 1, m, x);
          if (ind == -1) {
            cout << ind << endl;
          } else {
            v = a[ind];
            cout << dis[vec[ind - 1].S] << endl;
          }
        }
      }
    }

  return 0;
}