Codeforces question on Binary Lifting and LCA - c++

Question

Recently, I was solving this problem on Codeforces, but I failed the testcases. I can't see the testcases, nor can I see other's solutions because it is a gym question. Since, I cannot view the testcases where my code is failing, I need some help. I promise that the code is quite readable. Question is mainly on Binary lifting and LCA calculation.
As probably know, sloths live on trees. Accordingly, David has a pet sloth which he lets play on his unweighted trees when he solves programming problems. Occasionally, David will notice that his sloth is located on a particular node 𝑎 in the tree, and ask it to move to some other node 𝑏.
Of course, the sloth is as well-intentioned as can be, but alas, it only has enough energy to move across at most 𝑐 edges. If the sloth needs to cross fewer than 𝑐 edges to get to node 𝑏, it will get there and then take a nap. Otherwise, it will get as close as possible before it retires and hangs idly awaiting further digestion.
Where will the sloth end up? Also, since this happens quite often, David would like you to answer 𝑞 queries, each one of the similar form.
Input
The first line will contain a single integer 𝑛, the number of nodes in the tree. The following 𝑛−1 lines will contain two integers 𝑢 and 𝑣, describing the edges in the tree. These edges will form a tree.
After that, there will be a single line containing an integer 𝑞, the number of times David will motivate his sloth to move. 𝑞 lines follow, each containing three integers 𝑎, 𝑏, and 𝑐: the node the sloth starts on, the node David asks the sloth to move to, and the energy the sloth has when starting.
int LOGN = 19;
int n;
int timer = 0;
vector<int> adj[300005];
bool vis[300005];
int p[300005], tin[300005], tout[300005];
vector<int> d(300005,0);
int dp[300005][20];
void dfs(int v, int parent)
{
tin[v] = ++timer;
if (parent != -1)
d[v] = d[parent] + 1;
p[v] = parent;
dp[v][0] = parent;
for (int i = 1; i < LOGN; i++){
if (dp[v][i — 1] == -1)
dp[v][i] = -1;
else
dp[v][i] = dp[dp[v][i — 1]][i — 1];
}
for (auto u : adj[v]){
if (u != parent)
dfs(u, v);
}
tout[v] = ++timer;
}
bool isAncestor(int p, int c){ // check if p is ancestor of c
return (tin[p] <= tin[c] && tout[p] >= tout[c]);
}
int lift(int v, int steps) //find the y-th ancestor of x
{
for (int i = 0; i < 19; ++i){
if ((1 << i) & steps){
v = dp[v][i];
if (v == -1)
return -1;
steps -= (1 << i);
}
}
return v;
}
int LCA(int u, int v)
{
if (isAncestor(u, v))
return u;
if (isAncestor(v, u))
return v;
for (int i = 18; i >= 0; i--)
{
if (dp[u][i] == -1)// -1 should not be passed
continue;
if (!isAncestor(dp[u][i], v))
u = dp[u][i];
}
return dp[u][0];
}
void solve()
{
cin>>n;
timer=0;
for(int i=0;i<n-1;i++){// Input
int x,y;
cin>>x>>y;
x--;y--;
adj[x].pb(y);
adj[y].pb(x);
}
dfs(0,-1);
tout[0]=++timer;
int q;
cin>>q;
for(int z=0;z<q;z++){
int u,v,energy;
cin>>u>>v>>energy;
u--;v--;
if(isAncestor(u,v))// u is ancestor of v
{
if(d[v]-d[u]<=energy)// reachable
cout<<v+1<<endl;
else{// midway: lift
int res=lift(v,energy);
cout<<res+1<<endl;
}
}
else if(isAncestor(v,u))// v is ancestor of u
{
if(d[u]-d[v]<=energy)// reachable
cout<<v+1<<endl;
else{/// midway: lift
int res=lift(u,d[u]-d[v]-energy);
cout<<res+1<<endl;
}
}
else{
int lca=LCA(u,v);
int dist=d[u]+d[v]-2*d[lca];
if(energy>=dist)
cout<<v+1<<endl;
else{
// find vertex of the path from u to v with distance of energy from u
int dist1=d[u]-d[lca];
int dist2=d[v]-d[lca];
// lies on the path from u to LCA
if(energy<=dist1){
int res=lift(u,energy)+1;
cout<<res<<endl;
}
// lies on the path from LCA to v
else{
int rem=energy-dist1;
dist2-=rem;
int res=lift(v,dist2)+1;
cout<<res<<endl;
}
}
}
}
}
signed main()
{
FAST;
solve();
return 0;
}