geometry/Minkowski.h

• View this file on GitHub
• Last update: 2025-11-21 16:12:02+07:00

Depends on

• geometry/Point.h

Code

#include "Point.h"

vector<Point> MinkowskiSum(vector<Point> P, vector<Point> Q) {
  int n = P.size(), m = Q.size();
  vector<Point> R = {P[0] + Q[0]};
  for (int i = 1, j = 1; i < n || j < m; ) {
    if (i < n && (j == m || cross(P[i] - P[i - 1], Q[j] - Q[j - 1]) > 0)) {
      R.push_back(R.back() + P[i] - P[i - 1]);
      ++i;
    } else {
      R.push_back(R.back() + Q[j] - Q[j - 1]);
      ++j;
    }
  }
  return R;
}

#line 2 "geometry/Point.h"

template <class T>
int sgn(T x) { return (x > 0) - (x < 0); }
template <class T>
struct Point {
  typedef Point P;
  T x, y;
  explicit Point(T x = 0, T y = 0) : x(x), y(y) {}
  bool operator<(P p) const { return tie(x, y) < tie(p.x, p.y); }
  bool operator==(P p) const { return tie(x, y) == tie(p.x, p.y); }
  P operator+(P p) const { return P(x + p.x, y + p.y); }
  P operator-(P p) const { return P(x - p.x, y - p.y); }
  P operator*(T d) const { return P(x * d, y * d); }
  P operator/(T d) const { return P(x / d, y / d); }
  T dot(P p) const { return x * p.x + y * p.y; }
  T cross(P p) const { return x * p.y - y * p.x; }
  T cross(P a, P b) const { return (a - *this).cross(b - *this); }
  T dist2() const { return x * x + y * y; }
  T dist() const { return sqrt(dist2()); }
  // angle to x-axis in interval [-pi, pi]
  T angle() const { return atan2l(y, x); }
  P unit() const { return *this / dist(); }  // makes dist()=1
  P perp() const { return P(-y, x); }        // rotates +90 degrees
  P normal() const { return perp().unit(); }
  // returns point rotated 'a' radians ccw around the origin
  P rotate(ld a) const {
    return P(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
  }
  friend ostream& operator<<(ostream& os, P p) {
    return os << "(" << p.x << "," << p.y << ")";
  }
};
#line 2 "geometry/Minkowski.h"

vector<Point> MinkowskiSum(vector<Point> P, vector<Point> Q) {
  int n = P.size(), m = Q.size();
  vector<Point> R = {P[0] + Q[0]};
  for (int i = 1, j = 1; i < n || j < m; ) {
    if (i < n && (j == m || cross(P[i] - P[i - 1], Q[j] - Q[j - 1]) > 0)) {
      R.push_back(R.back() + P[i] - P[i - 1]);
      ++i;
    } else {
      R.push_back(R.back() + Q[j] - Q[j - 1]);
      ++j;
    }
  }
  return R;
}

Back to top page