HackerEarth Minimum Nodes problem solution

In this HackerEarth Minimum Nodes problem solution You are given a tree with N nodes and each node has some value assigned to it. Now you are given Q tasks of the form X K.

For each task, you have to find the path starting from X such that sum of nodes in the path is at least K and it contains minimum number of nodes. If such path exists, print the count of nodes in the path, else print -1.

#include<bits/stdc++.h>
#define ll long long
#define mp make_pair
#define pb push_back
#define si(x) scanf("%d",&x)
#define pi(x) printf("%dn",x)
#define s(x) scanf("%lld",&x)
#define p(x) printf("%lldn",x)
#define sc(x) scanf("%s",x)
#define pc(x) printf("%s",x)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define F first
#define S second
#define inf 4e18
#define prec(x) fixed<<setprecision(15)<<x
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define mem(x,y) memset(x,y,sizeof(x))
#define PQG priority_queue< int,std::vector<int>,std::greater<int> >
#define PQL priority_queue< int,std::vector<int>,std::less<int> >
#define PQPL priority_queue<pii ,vector< pii >, less< pii > >
#define PQPG priority_queue<pii ,vector< pii >, greater< pii > >
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)

using namespace std;

const int N=1e5+7;
vector<int>g[N];
ll a[N],k;
int ans,n,q;

void dfs(int node,int par,ll sum,int cnt) {
    sum+=a[node];
    cnt++;
    if(sum>=k) ans=min(ans,cnt);
    for(int i=0;i<g[node].size();i++) {
        int nxt=g[node][i];
        if(nxt==par) continue;
        dfs(nxt,node,sum,cnt);
    }
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("in09.txt","r",stdin);
        freopen("out09.txt","w",stdout);     
    #endif
    cin>>n>>q;
    assert(n>=1 && n<=100000);
    assert(q>=1 && q<=10);
    for(int i=1;i<=n;i++) {
        cin>>a[i];
        assert(a[i]>=1 && a[i]<=1000000000);
    }
    for(int i=1;i<n;i++) {
        int u,v;
        cin>>u>>v;
        assert(u>=1 && u<=n);
        assert(v>=1 && v<=n);
        g[u].pb(v);
        g[v].pb(u);
    }
    while(q--) {
        int src; 
        cin>>src>>k;
        ans=INT_MAX;
        dfs(src,0,0,0);
        if(ans==INT_MAX) ans=-1;
        cout<<ans<<endl;
    }
    
    return 0;
}