Lazy Salesgirl

PTA | 程序设计类实验辅助教学平台 ZOJ3606

枚举每个价格的数量，所以开3颗线段树维护即可

// Problem: G - Lazy Salesgirl
// Contest: Virtual Judge - The 9th Zhejiang Provincial Collegiate Programming Contest
// URL: https://vjudge.net/contest/16323#problem/G
// Memory Limit: 64 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstring>
#include<algorithm>
#include<iomanip>
#define INF (1<<30)
using namespace std;
typedef long long ll;
typedef double d;
const int N=1e5+9;
struct customer{
	int p,t;
	friend bool operator < (const customer &a,const customer &b){
		return a.t<b.t;
	}
}c[N];
struct inf{
	int id,w;
	friend bool operator < (const inf &a,const inf &b){
		return a.w<b.w;
	}
}pos[N];
int sum[N<<2];
struct node{
	d val;
}seg[4][N<<2];
ll tl(ll x){return x<<1;}
ll tr(ll x){return x<<1|1;}
void pushup(int id){
	sum[id]=sum[tl(id)]+sum[tr(id)];
	for(int i=1;i<=3;i++){
		int t=(i+sum[tl(id)]-1)%3+1;
		seg[i][id].val=seg[i][tl(id)].val+seg[t][tr(id)].val;
	}
}
void update(int id,int l,int r,int pos,ll v){
	if(l==r){
		sum[id]++;
		for(int i=1;i<=3;i++){
			seg[i][id].val=i*v;
		}
		return;
	}
	int mid=(l+r)>>1;
	if(mid>=pos){
		update(tl(id),l,mid,pos,v);
	}else{
		update(tr(id),mid+1,r,pos,v);
	}
	pushup(id);
}
void solve(){
	memset(sum,0,sizeof(sum));
	memset(seg,0,sizeof(seg));
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>c[i].p;
	}
	for(int i=1;i<=n;i++){
		cin>>c[i].t;
	}
	sort(c+1,c+1+n);
	pos[1].id=1;
	pos[1].w=c[1].t;
	for(int i=2;i<=n;i++){
		pos[i].id=i;
		pos[i].w=(c[i].t-c[i-1].t);
	}
	sort(pos+1,pos+1+n);
	d w=0;
	d ans=0;
	int j=0;
	for(int i=1;i<=n;){
		j=i;
		while(j<=n && pos[j].w==pos[i].w){
			update(1,1,n,pos[j].id,c[pos[j].id].p);
			j++;
		}
		d s=1.0*seg[1][1].val/sum[1];
		if(s>ans){
			ans=s;
			w=pos[i].w;
		}
		i=j;
	}
	cout<<fixed<<setprecision(6)<<w<<" "<<ans<<'\n';
}
int main(){
	int t;
	cin>>t;
	while(t--){
		solve();
	}
	return 0;
}