This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
struct Point {
double x, y;
};
/**
* @return the intersection of the perpendicular line that
* crosses the midpoint of Points a & b with the highway
*/
double max_point(const Point &a, const Point &b) {
return (b.x * b.x + b.y * b.y - a.x * a.x - a.y * a.y) /
(2 * b.x - 2 * a.x);
}
/** @return the euclidean distance between points a & b */
double dist(const Point &a, const Point &b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
int main() {
int n, l;
cin >> n >> l;
deque<Point> stck;
for (int i = 0; i < n; i++) {
Point cur;
cin >> cur.x >> cur.y;
cur.y = abs(cur.y);
// always take the one with the lowest y coord
if ((int)stck.size() && 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 ((int)stck.size() > 1 && max_point(stck[(int)stck.size() - 2], stck.back()) > max_point(stck.back(), cur)) {
stck.pop_back();
}
stck.push_back(cur);
}
// out of bounds
while ((int)stck.size() > 1 && max_point(stck[0], stck[1]) < 0) {
stck.pop_front();
}
while ((int)stck.size() > 1 && max_point(stck[(int)stck.size() - 2], stck.back()) > l) {
stck.pop_back();
}
double ans = 0;
for (int x = 0; x < (int)stck.size(); x++) {
// get critical points stck[x] is in charge of
Point left = {0, 0};
Point right {l, 0};
if (x) { left.x = max_point(stck[x], stck[x - 1]); }
if (x < (int)stck.size() - 1) { right.x = max_point(stck[x], stck[x + 1]); }
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;
}
Compilation message (stderr)
mobile.cpp: In function 'int main()':
mobile.cpp:61:16: warning: narrowing conversion of 'l' from 'int' to 'double' [-Wnarrowing]
61 | Point right {l, 0};
| ^
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