답안 #420491

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
420491 2021-06-08T11:58:55 Z palilo 두 개의 원 (balkan11_2circles) C++17
40 / 100
112 ms 4552 KB
#include <bits/stdc++.h>
using namespace std;

template <class T>
bool chmin(T& _old, T _new) { return _old > _new && (_old = _new, true); }
template <class T>
bool chmax(T& _old, T _new) { return _old < _new && (_old = _new, true); }

template <typename T>
struct point2D {
	T x, y;
	point2D() : x(0), y(0) {}
	point2D(T _x, T _y) : x(_x), y(_y) {}
	template <typename U>
	explicit point2D(const point2D<U>& p) : x(p.x), y(p.y) {}

	using P = point2D;

	bool operator<(const P& p) const { return tie(x, y) < tie(p.x, p.y); }
	bool operator==(const P& p) const { return tie(x, y) == tie(p.x, p.y); }
	bool operator!=(const P& p) const { return tie(x, y) != tie(p.x, p.y); }

	friend P operator+(const P& a, const P& b) { return P(a.x + b.x, a.y + b.y); }
	friend P operator-(const P& a, const P& b) { return P(a.x - b.x, a.y - b.y); }
	friend P operator*(const P& a, const T& scala) { return P(a.x * scala, a.y * scala); }
	friend P operator*(const T& scala, const P& a) { return P(scala * a.x, scala * a.y); }
	friend P operator/(const P& a, const T& scala) { return P(a.x / scala, a.y / scala); }

	friend ostream& operator<<(ostream& o, const P& p) { return o << '(' << p.x << ", " << p.y << ')'; }
	friend istream& operator>>(istream& i, P& p) { return i >> p.x >> p.y; }

	T dot(const P& p) const { return x * p.x + y * p.y; }
	T cross(const P& p) const { return x * p.y - y * p.x; }
	T cross(const P& a, const P& b) const { return (a - *this).cross(b - *this); }

	T dist2() const { return x * x + y * y; }
	double dist() const { return sqrt(dist2()); }

	P conj() const { return P(x, -y); }
	P perp_cw() const { return P(y, -x); }
	P perp_ccw() const { return P(-y, x); }

	P unit() const { return *this / dist(); }
	P normal() const { return perp_ccw().unit(); }

	P unit_int() const { return x || y ? *this / gcd(x, y) : *this; }
	P normal_int() const { return perp_ccw().unit_int(); }

	bool same_dir(const P& p) const { return cross(p) == 0 && dot(p) > 0; }
	bool on_segment(const P& s, const P& e) const {
		if constexpr (is_integral_v<T>)
			return cross(s, e) == 0 && (s - *this).dot(e - *this) <= 0;
		else
			return cross(s, e) == 0 && (s - *this).dot(e - *this) <= 1e-8;
	}
	int side_of(const P& s, const P& e) const {
		if constexpr (is_integral_v<T>) {
			auto c = s.cross(e, *this);
			return (c > 0) - (c < 0);
		} else {
			auto a = (e - s).cross(*this - s);
			double l = (e - s).dist() * 1e-8;
			return (a > l) - (a < -l);
		}
	}

	double angle() const { return atan2(y, x); }
	P rotate(double radian) const {
		return P(x * cos(radian) - y * sin(radian), x * sin(radian) + y * cos(radian));
	}
};

template <typename T>
double hull_diameter(const vector<point2D<T>>& hull) {
	T diameter = 0;
	int n = hull.size();
	for (int i = 0, j = n > 1; i < j; ++i) {
		for (;; j = j == n - 1 ? 0 : j + 1) {
			chmax(diameter, (hull[i] - hull[j]).dist2());
			if ((hull[(j + 1) % n] - hull[j]).cross(hull[i + 1] - hull[i]) >= 0)
				break;
		}
	}
	return sqrt(diameter);
}

template <class P = point2D<double>>
P segment_inter(const pair<P, P>& l1, const pair<P, P>& l2) {
	auto& [a, b] = l1;
	auto& [c, d] = l2;
	auto oa = c.cross(d, a), ob = c.cross(d, b);
	return (a * c.cross(d, b) - b * c.cross(d, a)) / (ob - oa);
}

int main() {
	cin.tie(nullptr)->sync_with_stdio(false);
#ifdef palilo
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
#endif

	using point = point2D<double>;

	int n;
	cin >> n;

	vector<point> a(n);
	for (auto& x : a) cin >> x;

	// precomputed values
	vector<point> perp_units(n);
	for (int i = 0; i < n - 1; ++i)
		perp_units[i] = (a[i + 1] - a[i]).perp_ccw().unit();
	perp_units[n - 1] = (a[0] - a[n - 1]).perp_ccw().unit();

	vector<pair<point, point>> edges;
	vector<point> mini_hull;
	edges.reserve(n), mini_hull.reserve(n);

	auto ok = [&](double R) -> bool {
		edges.clear();
		for (int i = 0; i < n; ++i) {
			pair cur = {a[i] + perp_units[i] * R, a[i == n - 1 ? 0 : i + 1] + perp_units[i] * R};
			// prev edge is outside the polygon
			while (!edges.empty() && edges.back().first.side_of(cur.first, cur.second) == -1)
				edges.pop_back();
			// prev edge is intersected
			if (!edges.empty()) {
				point inter = segment_inter(edges.back(), cur);
				cur.first = edges.back().second = inter;
			}
			edges.emplace_back(cur);
		}

		while (edges.size() > 1 && edges.back().first.side_of(edges.front().first, edges.front().second) == -1)
			edges.pop_back();
		if (edges.size() > 1) {
			point inter = segment_inter(edges.back(), edges.front());
			edges.front().first = edges.back().second = inter;
		}

		mini_hull.clear();
		for (const auto& [p, _] : edges) {
			if (!mini_hull.empty() && (p - mini_hull.back()).dist2() < 1e-8) continue;
			mini_hull.emplace_back(p);
		}
		if (mini_hull.size() > 1 && (mini_hull.front() - mini_hull.back()).dist2() < 1e-8)
			mini_hull.pop_back();
		return hull_diameter(mini_hull) >= 2 * R;
	};

	double lo = 0, hi = hull_diameter(a) / 4;
	while (hi - lo > 1e-4) {
		double mid = (lo + hi) / 2;
		(ok(mid) ? lo : hi) = mid;
	}

	cout << fixed << setprecision(3)
		 << lo;
}
# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 204 KB Output is correct
2 Incorrect 1 ms 204 KB Output isn't correct
3 Correct 2 ms 328 KB Output is correct
4 Correct 1 ms 332 KB Output is correct
5 Incorrect 5 ms 460 KB Output isn't correct
6 Incorrect 21 ms 1348 KB Output isn't correct
7 Incorrect 24 ms 1484 KB Output isn't correct
8 Incorrect 31 ms 1740 KB Output isn't correct
9 Incorrect 62 ms 3268 KB Output isn't correct
10 Correct 112 ms 4552 KB Output is correct