Submission #935281

#	Time	Username	Problem	Language	Result	Execution time	Memory
935281		cryan	Mobile (BOI12_mobile)	C++17	0 / 100	392 ms	48408 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

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));
	};

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

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

		if (idx > 1 && todo.back().x > max_point(stck[idx - 2], cur)) {
			idx--;
			todo.pop_back();
		}

		stck[idx] = cur;
		if (idx) {
			todo.push_back(Point{max_point(stck[idx - 1], cur), 0});
		}

		idx++;
	}

	double ans = 0;
	for (int x = 0; x < sz(todo); x++) {
		ans = max({ans, dist(todo[x], stck[x + 1]), dist(todo[x], stck[x])});
	}

	cout << 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.)

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