I have written this convex hull code which uses the atan() function from cmatch in order to calculate the polar angle of a point(angle of it's vector with Ox), then sort the points by this value. My solution is currently getting TLE and I would like to know if this is due to the atan function being slow. I have searched and have not been able to find any information on its performance. If atan is the cause I would appreciate suggestions on easy ways to circumvent this problem. I know there are ways to do convex hull without atan, it's just that calculating this is a pretty intuitive and easy way for me and I'm curious if there's any way to do it efficiently. Here is the code:
#include <bits/stdc++.h>
#define double long double
using namespace std;
ifstream in("infasuratoare.in");
ofstream out("infasuratoare.out");
struct Point {
double x,y;
bool operator < (const Point other) const {
if (y != other.y) {
return y < other.y;
}
return x < other.x;
}
Point operator - (const Point other) const {
return Point{x - other.x , y - other.y};
}
};
double angle(Point p) { /// returns in radians the angle of the Point
if (p.x == 0) {
if (p.y == 0) return 0;
if (p.y > 0) return M_PI/2;
if (p.y < 0) return M_PI + M_PI / 2;
}
if (p.x > 0 && p.y > 0) {
return atan(p.y / p.x);
} else if ( p.x > 0 && p.y < 0) {
return atan(p.y / p.x) + 2 * M_PI;
} else {
return atan(p.y / p.x) + M_PI;
}
}
bool rightTurn(Point a, Point b, Point c) {
return angle(b - a) > angle(c - b);
}
vector<Point> pt;
int n;
vector<Point> convexHull() {
vector<Point> res;
Point lowLeft = pt[1];
for (int i = 2; i <= n; i++) {
if (pt[i] < lowLeft) {
lowLeft = pt[i];
}
}
sort(pt.begin() + 1, pt.end() , [&](Point a, Point b) {
///we move everyone down by this low left point, then sort this increasing by it's angle.
return angle(a - lowLeft) < angle(b - lowLeft);
});
for (int i = 1; i <= n; i++) {
while(res.size() >= 2 && rightTurn(res[(int)res.size() - 2] , res[(int)res.size() - 1] , pt[i])) {
res.pop_back();
}
res.push_back(pt[i]);
}
return res;
}
int main() {
in >> n;
pt.resize(n + 1);
for (int i = 1; i <= n; i++) {
double x,y;
in >> x >> y;
pt[i] = {x,y};
}
auto hull = convexHull();
out << fixed << setprecision(12);
out << hull.size() << "\n";
for (auto k : hull) {
out << k.x << " " << k.y << "\n";
}
out << "\n";
}
TL & restraints please?
3 ≤ N ≤ 120.000 -1.000.000.000 ≤ Xi,Yi ≤ 1.000.000.000 No 3 points are on the same line. 0.5 seconds time limit (the judge is a bit slower than CF though.)
I might be very stupid, but why don't you use ios::sync_with_stdio(false); and cin.tie(nullptr); in this code?
Because this code uses fle input and output so I don't think that would do anything.
Don't read
double
directly fromistream
. Tracking where the notation of the number ends is extremely slow due to the complicated format. Instead, read it to a temporary string.I changed input to this and the speed didn't really change: string N; in >> N; n = stold(N);
string X,Y; in >> X >> Y; x = stold(X); y = stold(Y);
I also changed output to using to_string()
The problem with your code is probably with input like others said. However, convex hull code can be much shorter and without atan...