I was using the code to solve 258E - Little Elephant and Tree:

#include<cstdio>
#include<set>
#include<vector>
using namespace std;
int n,dfs[100009],dfscnt,ot[100009],ots[100009];
vector<int>e[100009];
void tree_srh(int nk,int l){
	dfs[nk]=++dfscnt;
	for(int i=0;i<e[nk].size();i++){
		if(e[nk][i]==l)continue; 
		tree_srh(e[nk][i],nk);
	}
	ot[nk]=ots[dfs[nk]]=dfscnt;
} 
vector<int>seg_tree[(1<<21)];
int m,a,b;
multiset<int>s;
int ans[100009];
void add(int k,int l,int r,const int lq,const int rq,const int n_val){
//printf("%d %d %d %d %d\n",k,l,r,lq,rq);
	if(l>rq||r<lq)return;
	if(l>=lq&&r<=rq){
		seg_tree[k].push_back(n_val); 
		return;
	}
	int m=(l+r)>>1;
	add((k<<1),l,m,lq,rq,n_val);
	add((k<<1)|1,m+1,r,lq,rq,n_val);
} 
int tamp(){
	int u=1,ats=0;
	while(u<=n){
		multiset<int>::iterator it=s.lower_bound(u);
		if(it==s.end())break;
		ats+=ots[*it]-*it+1;
		u=ots[*it]+1;
	}
	return ats;
}
void solve(int k,int l,int r){
	//printf("%d %d %d\n",k,l,r); 
	for(int i=0;i<seg_tree[k].size();i++)s.insert(seg_tree[k][i]);
	if(l==r){
		ans[l]=tamp();
		if(!s.empty())ans[l]--;
	} 
	else{
		int m=(l+r)>>1;
		solve((k<<1),l,m);
		solve((k<<1)|1,m+1,r);
	}
	for(int i=0;i<seg_tree[k].size();i++)s.erase(s.find(seg_tree[k][i]));
}
int main(){
	scanf("%d %d",&n,&m);
	for(int i=1;i<n;i++){
		scanf("%d %d",&a,&b);
		e[a].push_back(b);
		e[b].push_back(a);
	}
	tree_srh(1,0);
	/*for(int i=1;i<=n;i++){
		printf("%d ",dfs[i]);
	}
	putchar('\n');*/
	for(int i=1;i<=m;i++){
		scanf("%d %d",&a,&b);
		add(1,1,n,dfs[a],ot[a],dfs[a]);
		add(1,1,n,dfs[a],ot[a],dfs[b]);
		add(1,1,n,dfs[b],ot[b],dfs[a]);
		add(1,1,n,dfs[b],ot[b],dfs[b]);
	}
	solve(1,1,n);
	for(int i=1;i<=n;i++){
		printf("%d ",ans[dfs[i]]);
	}
	return 0;
}

The code passed the task quickly but I couldn't prove it was at the right time complexy.Could anyone tell me how to prove the time complexy of the code that is enough to pass the task or hack it? waiting for your answer.