2
5 10
20 25

10 30

区间选取

用Past和In_s维护经过的左右结点，

并由此算出

嵌套数（左-右）

左边的区间数=右

右边的区间数=总-（左-右）-右=总-左

PS:由于是闭区间，需要把-右的时间后延

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<iostream>
#include<functional>
#include<algorithm>
using namespace std;
#define MAXN (200000+10)
#define MAXAi (1000000000)
#define NDEBUG
struct segment_node
{
	int l,type;
	friend bool operator<(const segment_node a,const segment_node b){return (a.l!=b.l)?a.l<b.l:a.type<b.type;	}
	friend bool operator==(const segment_node a,const segment_node b){return (a.l==b.l&&a.type==b.type);	}
	friend bool operator!=(const segment_node a,const segment_node b){return (!(a==b));	}

}a[MAXN];
int n;
long long next[MAXN];
int main()
{
	#ifndef NDEBUG
	freopen("segmentbet.in","r",stdin);
	#endif
	memset(a,0,sizeof(a));
	scanf("%d",&n);n<<=1;
	for (int i=1;i<=n;i+=2)
	{
		scanf("%d%d",&a[i].l,&a[i+1].l);
		a[i].type=-1;a[i+1].type=1;
	}
	sort(a+1,a+n+1);
	int in_s=0,minE=0;
	long long ans=0;
	memset(next,0,sizeof(next));
	for (int i=n-1;i>=1;i--)
	{
		next[i]=next[i+1];
		if (a[i+1].type==-1)
			next[i]+=a[i+1].l;
	}
	int j=1,past=0,pastE=0;
	for (int i=1;i<=n;i++)
	{
		if (a[i]!=a[i+1])
		{
			int len=i-j+1;
			j=i+1;
			len*=-a[i].type;
			if (pastE!=a[i].l) {in_s+=past;past=0;}
			if (len>0) in_s+=len;
			else past+=len;

			long long cost=(long long)(long long)in_s*(long long)a[i].l+next[i];
			if (ans<cost)
			{
				ans=cost;
				minE=a[i].l;
			}

		}

	}
	cout<<minE<<' '<<ans<<endl;
	return 0;
}