A. Access Points(单调栈&贪心)

Posted 2021-12-23 Harris-H

A. Access Points(单调栈&贪心)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e5+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n)
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\\n",a[n]); 

template <typename T>		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y)
	if(x<y) x=y;

template <typename T>
void cmn(T &x,T y)
	if(x>y) x=y;

int n;
double solve(int a[])
	stack<pair<ll,int> >s;
	rep(i,1,n)
		ll x=a[i],y=1;
		while(!s.empty()&&s.top().x*y>=s.top().y*x)
			x+=s.top().x;
			y+=s.top().y;
			s.pop();
		
		s.push(x,y);
	
	int x=n;
	double ans=0;
	while(!s.empty())
		double val = s.top().x;ll cnt=s.top().y;
		val/=cnt;
		s.pop();
		for(int i=x-cnt+1;i<=x;i++)
			ans+=(a[i]-val)*(a[i]-val);
		
		x-=cnt;
	
	return ans;

int a[N],b[N];
int main()
	scanf("%d",&n);
	rep(i,1,n) scanf("%d%d",&a[i],&b[i]);
	printf("%f\\n",solve(a)+solve(b));
	return 0;