Submission #935546

#	Time	Username	Problem	Language	Result	Execution time	Memory
935546		cryan	Mobile (BOI12_mobile)	C++17	100 / 100	443 ms	33184 KiB

mobile

// oh, these hills, they burn so bright / oh, these hills, they bring me life
#include "bits/stdc++.h"
using namespace std;

using ll = long long;
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)(x.size())
#define inf 1000000010
#define linf 0x3f3f3f3f3f3f3f3f
#define mp make_pair
#define f first
#define s second
#define pi pair<int, int>
#ifdef LOCAL
#include "/mnt/c/yukon/pp.hpp"
#else
#define endl '\n'
#endif

#define double long double
struct Point {
	double x, y;
};

int main() {
	cin.tie(0)->sync_with_stdio(0);

	int n, l;
	cin >> n >> l;

	function<double(Point, Point)> max_point = [&](Point a, Point b) {
		// return (((b.x) * (b.x)) + ((b.y) * (b.y)) - ((a.x) * (a.x)) - ((a.y) * (a.y))) / (2 * (b.x - a.x));
		return (b.x * b.x + b.y * b.y - a.x * a.x - a.y * a.y) / (2 * b.x - 2 * a.x);
	};

	function<double(Point, Point)> dist = [&](Point a, Point b) {
		return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
	};

	deque<Point> stck;
	for (int i = 0; i < n; i++) {
		Point cur;
		cin >> cur.x >> cur.y;
		cur.y = abs(cur.y);

		// take the one with the lowest y coord
		if (sz(stck) && stck.back().x == cur.x) {
			if (cur.y >= stck.back().y) {
				continue;
			} else if (cur.y < stck.back().y) {
				stck.pop_back();
			}
		}

		// find "crosses"
		while (sz(stck) > 1 && max_point(stck[sz(stck) - 2], stck.back()) > max_point(stck.back(), cur)) {
			stck.pop_back();
		}

		stck.push_back(cur);
	}

	while (sz(stck) > 1 && max_point(stck[0], stck[1]) < 0)
		stck.pop_front();
	while (sz(stck) > 1 && max_point(stck[sz(stck) - 2], stck.back()) > l)
		stck.pop_back();

	double ans = 0;
	for (int x = 0; x < sz(stck); x++) {
		Point left = {(x ? max_point(stck[x], stck[x - 1]) : 0), 0};
		Point right = {(x < sz(stck) - 1 ? max_point(stck[x], stck[x + 1]) : l), 0};
		if (left.x < 0 || right.x > l || right.x < 0 || left.x > l)
			continue;

		ans = max({ans, dist(stck[x], left), dist(stck[x], right)});
	}

	cout << fixed << setprecision(6) << ans << endl;
}

// don't get stuck on one approach
// question bounds
// flesh out your approach before implementing o.o
// math it out
// ok well X is always possible, how about X + 1 (etc.)

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 4 / 4
• Subtask 4: 4 / 4
• Subtask 5: 4 / 4
• Subtask 6: 5 / 5
• Subtask 7: 5 / 5
• Subtask 8: 5 / 5
• Subtask 9: 5 / 5
• Subtask 10: 5 / 5
• Subtask 11: 5 / 5
• Subtask 12: 5 / 5
• Subtask 13: 5 / 5
• Subtask 14: 5 / 5
• Subtask 15: 5 / 5
• Subtask 16: 5 / 5
• Subtask 17: 5 / 5
• Subtask 18: 5 / 5
• Subtask 19: 5 / 5
• Subtask 20: 5 / 5
• Subtask 21: 5 / 5