Sorting out interval greed problems

I've been studying for sevenoreight months , Although I did brush some questions , But the strength is still the same lj, Some of the topics written at that time were not well sorted out , There are many classic topics that haven't gone deep , How to improve ?

Problem 905. Interval selection

brush 112. Radar equipment We need to use the greedy strategy of selecting points in the interval , At that time, I remember that most of the greedy strategies about interval problems in the basic course were sorted by the right endpoint , When learning, I don't quite understand why I don't use the left endpoint to sort , After seven 、 When I write again in eight months , I found a template problem is not very clear , It's time to be beaten

$It\text{'}s\; written\; in\; left\; endpoint\; sort\; ,\; When\; analyzing\; later,\; it\; is\; obvious\; that\; there\; are\; counter\; examples,2/10(Only\; two\; test\; points\; can\; be\; passed)$

 Counter example 	
3
1 4
2 6
3 5  The answer is 2

/*Love coding and thinking!*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#define pb push_back 
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo2(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
typedef pair<int,int>PII;
typedef pair<long,long>PLL;

typedef long long ll;
const int N=2e5+10;
int n,m,_,x;
int a[N],c[N];
int res=1,aver;

struct node
{
    
	int l,r;
}Node[N];

bool cmp(node a,node b)
{
    
	return a.l<b.l;
}

int main()
{
    
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d%d",&Node[i].l,&Node[i].r);
	
	sort(Node+1,Node+1+n,cmp);// Default left endpoint 
	int last=Node[1].r;
	
	for(int i=2;i<=n;i++)
	{
    
		if(Node[i].l>Node[i-1].r)
			res++,last=max(last,Node[i].r);
		else
		{
    
			last=max(last,Node[i].r);
		}
	}
	cout<<res;
	return 0;
}

$Later,\; it\; was\; changed\; blindly\; in\; view\; of\; the\; above\; situation,emmm,It\; \text{'}s\; a\; long\; story,obviousn=1,Output0$

/*Love coding and thinking!*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#define pb push_back 
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo2(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
typedef pair<int,int>PII;
typedef pair<long,long>PLL;


typedef long long ll;
const int N=2e5+10;
int n,m,_,x;
int a[N],c[N];
int res=1,aver;

struct node
{
    
	int l,r;
}Node[N];

bool cmp(node a,node b)
{
    
	return a.l<b.l;
}

int main()
{
    
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d%d",&Node[i].l,&Node[i].r);
	
	sort(Node+1,Node+1+n,cmp);// Default left endpoint 
	int last=Node[1].r;s
	
	for(int i=2;i<=n;i++)
	{
       
		if(Node[i].l>Node[i-1].r)
			res++,last=max(last,Node[i].r);
		else
		{
    
			res++;
			last=Node[i].r;
		}
	}
	cout<<res-1;
	return 0;
}

$sortThere\; is\; still\; a\; problem\; with\; the\; right\; endpoint,There\text{'}s\; no\; one\; left$

/*Love coding and thinking!*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#define pb push_back 
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo2(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
typedef pair<int,int>PII;
typedef pair<long,long>PLL;
typedef long long ll;
const int N=2e5+10;

struct node
{
    
	int l,r;
}Node[N];

bool cmp(node a,node b)
{
    
	return a.r<b.r;
}
int main()
{
    
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d%d",&Node[i].l,&Node[i].r);
	
	sort(Node+1,Node+1+n,cmp);// Default left endpoint 
	
	int last=Node[1].r;
	
	for(int i=2;i<=n;i++)
	{
    
		if(Node[i].l>last)
			res++,last=Node[i].r;
		else
		{
    
			last=Node[i-1].r;
		}
	}
	cout<<res;
	return 0;
}

I'm a fool

/*Love coding and thinking!*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#define pb push_back 
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo2(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
typedef pair<int,int>PII;
typedef pair<long,long>PLL;
typedef long long ll;
const int N=2e5+10;

struct node
{
    
	int l,r;
}Node[N];
int res=1,n;
bool cmp(node a,node b)
{
    
	return a.r<b.r;
}
int main()
{
    
	cin>>n;
	for(int i=1;i<=n;i++)
		scanf("%d%d",&Node[i].l,&Node[i].r);
	
	sort(Node+1,Node+1+n,cmp);// Default left endpoint 
	// Greedy strategy , If the first i It's an interval l> The right end of the previous interval , answer ++
    // Otherwise, the right endpoint is updated .
	int last=Node[1].r;
	
	for(int i=2;i<=n;i++)
	{
    
		if(Node[i].l>last)
			res++,last=Node[i].r;
	}
	cout<<res;
	return 0;
}

Radar equipment

/*Love coding and thinking!*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#define pb push_back 
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo2(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
typedef pair<int,int>PII;
typedef pair<long,long>PLL;
typedef long long ll;
const int N=2e5+10;

struct node
{
    
	int l,r;
}Node[N];

int cnt=0;
struct seg
{
    
	double l,r;
}Seg[N];

bool cmp(seg a,seg b)
{
    
	return a.r<b.r;
}

int res=1,n,D;


int main()
{
    
	cin>>n>>D;
	for(int i=1;i<=n;i++)
		scanf("%d%d",&Node[i].l,&Node[i].r);
	
	for(int i=1;i<=n;i++)
	{
    
		if(Node[i].r>D)
		{
    
			cout<<"-1";
			return 0;		
		}
		else
		{
    
			double d=sqrt(D*D-Node[i].r*Node[i].r);
			Seg[cnt].l=Node[i].l-d;
			Seg[cnt++].r=Node[i].l+d;
		}
	}	
	
	sort(Seg,Seg+cnt,cmp);// Default left endpoint 
	
	// for(int i=0;i<cnt;i++)
	// {
    
		// cout<<Seg[i].l<<" "<<Seg[i].r<<endl;
	// }
	
	// Greedy strategy , If the first i It's an interval l> The right end of the previous interval , answer ++
    // Otherwise, the right endpoint is updated .
	double last=Seg[0].r;
	
	for(int i=1;i<cnt;i++)
	{
    
		if(Seg[i].l>last)
			res++,last=Seg[i].r;
	}
	cout<<res;
	return 0;
}