제출 #1304831

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1304831		neonglitch	Balloons (CEOI11_bal)	C++20	100 / 100	331 ms	11096 KiB

bal

#include <iostream>
#include <iomanip>
#include <cmath>
#include <vector>
using namespace std;
typedef long double ld;
typedef long double ll;
const int N=2e5+10;
ld x[N],r[N];
template<typename T> bool chmin(T& a, const T& b) {return b < a ? a = b, 1 : 0;}
template<typename T> bool chmax(T& a, const T& b) {return b > a ? a = b, 1 : 0;}
template<typename X, typename Y>
struct lichao {

	static constexpr Y inf = numeric_limits<Y>::max();

	struct Node {
		Node *l = 0, *r = 0;
		Y k = 0, b = inf;

		Node() = default;
	};

	X L, R;
	Node *root = new Node();

	lichao(X L, X R): L(L), R(R) {}

	void add_seg(X ql, X qr, X l, X r, Y k, Y b, Node*& n) {
		if (qr < l || r < ql) return;
		if (!n) n = new Node();
		X md = l + (r - l) / 2;
		if (l < ql || r > qr) {
			add_seg(ql, qr, l, md, k, b, n->l);
			add_seg(ql, qr, md + 1, r, k, b, n->r);
			return;
		}
		add_line(l, r, k, b, n);
	}

	void add_line(X l, X r, Y k, Y b, Node*& n) {
		if (!n) n = new Node();
		Y vl_cur = n->k * l + n->b;
		Y vr_cur = n->k * r + n->b;
		Y vl_new = k * l + b;
		Y vr_new = k * r + b;
		if ((vl_cur <= vl_new) == (vr_cur <= vr_new)) {
			if (vl_new < vl_cur) n->k = k, n->b = b;
			return;
		}
		X md = l + (r - l) / 2;
		Y vmd_cur = n->k * md + n->b;
		Y vmd_new = k * md + b;
		if (vmd_new < vmd_cur) {
			swap(n->k, k), swap(n->b, b);
			swap(vl_cur, vl_new);
		}
		if (vl_new < vl_cur) add_line(l, md, k, b, n->l);
		else add_line(md + 1, r, k, b, n->r);
	}

	Y get_min(X x) {
		Node *n = root;
		X l = L, r = R;
		Y o = inf;
		while (n) {
			chmin(o, n->k * x + n->b);
			X md = l + (r - l) / 2;
			(x <= md ? r : l) = md + (x > md);
			n = x <= md ? n->l : n->r;
		}
		return o;
	}

	void add_line(Y k, Y b) {add_line(L, R, k, b, root);}
	void add_seg(X ql, X qr, Y k, Y b) {add_seg(ql, qr, L, R, k, b, root);}
};
//add_line - O(log(C))
//add_seg - O(log(C) ^ 2)
//Usage: lichao<X, Y> kek(L, R), where
//X - type of coordinates,
//Y - type of K and B in linear function y = Kx + B,
//L, R - bounds of X.

const ll G = 1ll << 30;
lichao<ll, ll> kek(0, G);
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>x[i]>>r[i];
		if(i)
		{
			ll p=kek.get_min(x[i]);
			// cout<<"Hell "<<p<<endl;
			r[i]=min(r[i],p*p);
		}
		ld sr=sqrtl(4*r[i]);
		ld inv=1.0;
		inv/=sr;
		kek.add_line(inv,-x[i]/sr);
		// cout<<"add "<<inv<<' '<<(-x[i]/sr)<<endl;;
		cout<<fixed<<setprecision(3)<<r[i]<<endl;
	}
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10