Package delivery (greedy)

Portal
The question ： Next n Express delivery arrived one after another in the middle of the day , The delivery arrival time and the latest retrieval time are known , Every time k Express delivery , Ask at least how many times to take all the express
Greedy thoughts , Try to pick it up late , stay a[i].r Pick up
Sort each express according to the time of arrival , And, in turn, now comparison , stay now Previously arrived revenue queue
Be careful now Constantly updating , First, take the first arrival deadline , Of items arriving again r To update now（ I can't think of this ）
now From the queue a[i].r take (r Small before shooting ), Finished at the deadline of other goods Delete from the queue
If all the items in the queue are taken at one time , Just continue to traverse the items coming in later

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std; 
//#define int long long int
#define pii pair<int,int>
const int N=1e5+5;
struct node{
    
	int l,r;
	bool operator<(const node&b)const{
    
		return l<b.l;// Sort by arrival time , And, in turn, now comparison  
	} 
}a[N];
signed main(){
     
	int t,n,k,l,r;
	cin>>t;
	while(t--){
    // How many times at least  

// cin>>n>>k;
		scanf("%d %d",&n,&k);	
		for(int i=0;i<n;i++){
    
// cin>>a[i].l>>a[i].r; 
			scanf("%d %d",&a[i].l,&a[i].r);	
		}
		sort(a,a+n);
		priority_queue<pii,vector<pii>,greater<pii> > Q;
		while(!Q.empty())Q.pop();
		Q.push({
    a[0].r,a[0].l});
		int res=0;
		int now=a[0].r;
		int i=1;
		while(i<n){
    // Traverse the express delivered successively  
			while(i<n&&a[i].l<=now) {
    //now You must go and get it once , See what else can be taken  
				now=min(now,a[i].r);//now You should also find the first expired one in the express delivery that arrives  
				Q.push({
    a[i].r,a[i].l});
				i++;// Continue to look at the next  
			}
// if(!Q.empty()){// It can't be empty  
			res++;
			int s=Q.size();
			if(s<=k){
    
				while(!Q.empty())Q.pop();
				if(i<n){
    
					now=a[i].r;
					Q.push({
    a[i].r,a[i].l});
					i++; 
				}
			} 
			else{
    
				int c=k;
				while(c--)Q.pop();// Put the first to the deadline k Take it away , Otherwise, it will expire 
				now=Q.top().first; 
			}
		}
		res+=(Q.size()+k-1)/k;// Rounding up , Maybe the last time you can take away more than k 
		cout<<res<<endl; 
	}
	return 0;
}