CF845C Two TVs

https://www.luogu.com.cn/problem/CF845C
差分+离散化

#include<bits/stdc++.h>
using namespace std;
int a[700100],s[700100],se[700100],e[700100];//2*n的范围
int main()
{
	int n,sum=0;
	cin>>n;
	for(int i=1;i<=n;i++) 
	{
		cin>>s[i]>>e[i];
		se[2*i-1]=s[i],se[2*i]=e[i];
	}
	sort(se+1,se+1+2*n);
	int ms=unique(se+1,se+1+2*n)-1-se;
	for(int i=1;i<=n;i++)
	{
		s[i]=lower_bound(se+1,se+1+ms,s[i])-se;
		e[i]=lower_bound(se+1,se+1+ms,e[i])-se;
	}
	for(int i=1;i<=n;i++) 
	{
		a[s[i]]++;
		a[e[i]+1]--;
	}
	for(int i=0;i<=2*n;i++) //可能从零开始
	{
		sum+=a[i];
		if(sum>2) 
		{
			cout<<"NO";
			return 0;
		}
	}
	cout<<"YES";
	return 0;
}```