geo/convex_polygon.hpp

• View this file on GitHub
• Last update: 2025-05-05 02:10:07+09:00
• Include: #include "geo/convex_polygon.hpp"
• Link: https://codeforces.com/contest/1906/problem/D
• Link: https://codeforces.com/contest/799/problem/G

Depends on

• geo/base.hpp
• geo/convex_hull.hpp

Required by

• geo/minkowski_sum.hpp

Verified with

• test/1_mytest/convex_polygon_side.test.cpp
• test/1_mytest/convex_polygon_visible_range.test.cpp
• test/1_mytest/max_dot.test.cpp

Code

#pragma once

#include "geo/base.hpp"
#include "geo/convex_hull.hpp"

// n=2 は現状サポートしていない
template <typename T>
struct ConvexPolygon {
  using P = Point<T>;
  int n;
  vc<P> point;
  T area2;

  ConvexPolygon(vc<P> point_) : n(len(point_)), point(point_) {
    assert(n >= 3);
    area2 = 0;
    FOR(i, n) {
      int j = nxt_idx(i), k = nxt_idx(j);
      assert((point[j] - point[i]).det(point[k] - point[i]) >= 0);
      area2 += point[i].det(point[j]);
    }
  }

  // 比較関数 comp(i,j)
  template <typename F>
  int periodic_min_comp(F comp) {
    int L = 0, M = n, R = n + n;
    while (1) {
      if (R - L == 2) break;
      int L1 = (L + M) / 2, R1 = (M + R + 1) / 2;
      if (comp(L1 % n, M % n)) { R = M, M = L1; }
      elif (comp(R1 % n, M % n)) { L = M, M = R1; }
      else {
        L = L1, R = R1;
      }
    }
    return M % n;
  }

  int nxt_idx(int i) { return (i + 1 == n ? 0 : i + 1); }
  int prev_idx(int i) { return (i == 0 ? n - 1 : i - 1); }

  // 中：1, 境界：0, 外：-1. test した.
  int side(P p) {
    int L = 1, R = n - 1;
    T a = (point[L] - point[0]).det(p - point[0]);
    T b = (point[R] - point[0]).det(p - point[0]);
    if (a < 0 || b > 0) return -1;
    // p は 0 から見て [L,R] 方向
    while (R - L >= 2) {
      int M = (L + R) / 2;
      T c = (point[M] - point[0]).det(p - point[0]);
      if (c < 0)
        R = M, b = c;
      else
        L = M, a = c;
    }
    T c = (point[R] - point[L]).det(p - point[L]);
    T x = min({a, -b, c});
    if (x < 0) return -1;
    if (x > 0) return 1;
    // on triangle p[0]p[L]p[R]
    if (p == point[0]) return 0;
    if (c != 0 && a == 0 && L != 1) return 1;
    if (c != 0 && b == 0 && R != n - 1) return 1;
    return 0;
  }

  // return {min, idx}. test した.
  pair<T, int> min_dot(P p) {
    int idx = periodic_min_comp([&](int i, int j) -> bool { return point[i].dot(p) < point[j].dot(p); });
    return {point[idx].dot(p), idx};
  }

  // return {max, idx}. test した.
  pair<T, int> max_dot(P p) {
    int idx = periodic_min_comp([&](int i, int j) -> bool { return point[i].dot(p) > point[j].dot(p); });
    return {point[idx].dot(p), idx};
  }

  // p から見える範囲. p 辺に沿って見えるところも見えるとする. test した.
  // 多角形からの反時計順は [l,r] だが p から見た偏角順は [r,l] なので注意
  pair<int, int> visible_range(P p) {
    int a = periodic_min_comp([&](int i, int j) -> bool { return ((point[i] - p).det(point[j] - p) < 0); });
    int b = periodic_min_comp([&](int i, int j) -> bool { return ((point[i] - p).det(point[j] - p) > 0); });
    if ((p - point[a]).det(p - point[prev_idx(a)]) == T(0)) a = prev_idx(a);
    if ((p - point[b]).det(p - point[nxt_idx(b)]) == T(0)) b = nxt_idx(b);
    return {a, b};
  }

  // 線分が「内部と」交わるか
  // https://codeforces.com/contest/1906/problem/D
  bool check_cross(P A, P B) {
    FOR(2) {
      swap(A, B);
      auto [a, b] = visible_range(A);
      if ((point[a] - A).det(B - A) >= 0) return 0;
      if ((point[b] - A).det(B - A) <= 0) return 0;
    }
    return 1;
  }

  vc<T> AREA;

  // point[i,...,j] (inclusive) の面積の 2 倍
  T area_between(int i, int j) {
    assert(i <= j && j <= i + n);
    if (j == i + n) return area2;
    i %= n, j %= n;
    if (i > j) j += n;
    if (AREA.empty()) build_AREA();
    return AREA[j] - AREA[i] + (point[j % n].det(point[i]));
  }

  void build_AREA() {
    AREA.resize(2 * n);
    FOR(i, n) AREA[n + i] = AREA[i] = point[i].det(point[nxt_idx(i)]);
    AREA = cumsum<T>(AREA);
  }

  // 直線の左側の面積. strict に 2 回交わることを仮定.
  // https://codeforces.com/contest/799/problem/G
  T left_area(Line<T> L) {
    static_assert(is_same<T, double>::value || is_same<T, long double>::value);
    Point<T> normal(L.a, L.b);
    int a = min_dot(normal).se;
    int b = max_dot(normal).se;
    if (b < a) b += n;
    assert(L.eval(point[a % n]) < 0 && L.eval(point[b % n]) > 0);
    int p = binary_search([&](int i) -> bool { return L.eval(point[i % n]) < 0; }, a, b);
    int q = binary_search([&](int i) -> bool { return L.eval(point[i % n]) > 0; }, b, a + n);
    T s, t;
    {
      T x = L.eval(point[p % n]);
      T y = L.eval(point[(p + 1) % n]);
      s = x / (x - y);
    }
    {
      T x = L.eval(point[q % n]);
      T y = L.eval(point[(q + 1) % n]);
      t = x / (x - y);
    }
    P A(point[p % n]), B(point[(p + 1) % n]);
    P C(point[q % n]), D(point[(q + 1) % n]);
    P X = B * s + A * (1 - s);
    P Y = D * t + C * (1 - t);
    T ANS = area_between(p, q);
    ANS -= (A - C).det(X - C);
    ANS += (Y - C).det(X - C);
    return ANS;
  }
};

#line 2 "geo/convex_polygon.hpp"

#line 2 "geo/base.hpp"
template <typename T>
struct Point {
  T x, y;

  Point() : x(0), y(0) {}

  template <typename A, typename B>
  Point(A x, B y) : x(x), y(y) {}

  template <typename A, typename B>
  Point(pair<A, B> p) : x(p.fi), y(p.se) {}

  Point operator+=(const Point p) {
    x += p.x, y += p.y;
    return *this;
  }
  Point operator-=(const Point p) {
    x -= p.x, y -= p.y;
    return *this;
  }
  Point operator+(Point p) const { return {x + p.x, y + p.y}; }
  Point operator-(Point p) const { return {x - p.x, y - p.y}; }
  bool operator==(Point p) const { return x == p.x && y == p.y; }
  bool operator!=(Point p) const { return x != p.x || y != p.y; }
  Point operator-() const { return {-x, -y}; }
  Point operator*(T t) const { return {x * t, y * t}; }
  Point operator/(T t) const { return {x / t, y / t}; }

  bool operator<(Point p) const {
    if (x != p.x) return x < p.x;
    return y < p.y;
  }
  T dot(const Point& other) const { return x * other.x + y * other.y; }
  T det(const Point& other) const { return x * other.y - y * other.x; }

  double norm() { return sqrtl(x * x + y * y); }
  double angle() { return atan2(y, x); }

  Point rotate(double theta) {
    static_assert(!is_integral<T>::value);
    double c = cos(theta), s = sin(theta);
    return Point{c * x - s * y, s * x + c * y};
  }
  Point rot90(bool ccw) { return (ccw ? Point{-y, x} : Point{y, -x}); }
};

#ifdef FASTIO
template <typename T>
void rd(Point<T>& p) {
  fastio::rd(p.x), fastio::rd(p.y);
}
template <typename T>
void wt(Point<T>& p) {
  fastio::wt(p.x);
  fastio::wt(' ');
  fastio::wt(p.y);
}
#endif

// A -> B -> C と進むときに、左に曲がるならば +1、右に曲がるならば -1
template <typename T>
int ccw(Point<T> A, Point<T> B, Point<T> C) {
  T x = (B - A).det(C - A);
  if (x > 0) return 1;
  if (x < 0) return -1;
  return 0;
}

template <typename REAL, typename T, typename U>
REAL dist(Point<T> A, Point<U> B) {
  REAL dx = REAL(A.x) - REAL(B.x);
  REAL dy = REAL(A.y) - REAL(B.y);
  return sqrt(dx * dx + dy * dy);
}

// ax+by+c
template <typename T>
struct Line {
  T a, b, c;

  Line(T a, T b, T c) : a(a), b(b), c(c) {}
  Line(Point<T> A, Point<T> B) { a = A.y - B.y, b = B.x - A.x, c = A.x * B.y - A.y * B.x; }
  Line(T x1, T y1, T x2, T y2) : Line(Point<T>(x1, y1), Point<T>(x2, y2)) {}

  template <typename U>
  U eval(Point<U> P) {
    return a * P.x + b * P.y + c;
  }

  template <typename U>
  T eval(U x, U y) {
    return a * x + b * y + c;
  }

  // 同じ直線が同じ a,b,c で表現されるようにする
  void normalize() {
    static_assert(is_same_v<T, int> || is_same_v<T, long long>);
    T g = gcd(gcd(abs(a), abs(b)), abs(c));
    a /= g, b /= g, c /= g;
    if (b < 0) { a = -a, b = -b, c = -c; }
    if (b == 0 && a < 0) { a = -a, b = -b, c = -c; }
  }

  bool is_parallel(Line other) { return a * other.b - b * other.a == 0; }
  bool is_orthogonal(Line other) { return a * other.a + b * other.b == 0; }
};

template <typename T>
struct Segment {
  Point<T> A, B;

  Segment(Point<T> A, Point<T> B) : A(A), B(B) {}
  Segment(T x1, T y1, T x2, T y2) : Segment(Point<T>(x1, y1), Point<T>(x2, y2)) {}

  bool contain(Point<T> C) {
    T det = (C - A).det(B - A);
    if (det != 0) return 0;
    return (C - A).dot(B - A) >= 0 && (C - B).dot(A - B) >= 0;
  }

  Line<T> to_Line() { return Line(A, B); }
};

template <typename REAL>
struct Circle {
  Point<REAL> O;
  REAL r;
  Circle() {}
  Circle(Point<REAL> O, REAL r) : O(O), r(r) {}
  Circle(REAL x, REAL y, REAL r) : O(x, y), r(r) {}
  template <typename T>
  bool contain(Point<T> p) {
    REAL dx = p.x - O.x, dy = p.y - O.y;
    return dx * dx + dy * dy <= r * r;
  }
};
#line 2 "geo/convex_hull.hpp"

#line 4 "geo/convex_hull.hpp"

// allow_180=true で同一座標点があるとこわれる
// full なら I[0] が sorted で min になる
template <typename T, bool allow_180 = false>
vector<int> ConvexHull(vector<Point<T>>& XY, string mode = "full", bool sorted = false) {
  assert(mode == "full" || mode == "lower" || mode == "upper");
  ll N = XY.size();
  if (N == 1) return {0};
  if (N == 2) {
    if (XY[0] < XY[1]) return {0, 1};
    if (XY[1] < XY[0]) return {1, 0};
    return {0};
  }
  vc<int> I(N);
  if (sorted) {
    FOR(i, N) I[i] = i;
  } else {
    I = argsort(XY);
  }
  if constexpr (allow_180) { FOR(i, N - 1) assert(XY[i] != XY[i + 1]); }

  auto check = [&](ll i, ll j, ll k) -> bool {
    T det = (XY[j] - XY[i]).det(XY[k] - XY[i]);
    if constexpr (allow_180) return det >= 0;
    return det > T(0);
  };

  auto calc = [&]() {
    vector<int> P;
    for (auto&& k: I) {
      while (P.size() > 1) {
        auto i = P[P.size() - 2];
        auto j = P[P.size() - 1];
        if (check(i, j, k)) break;
        P.pop_back();
      }
      P.eb(k);
    }
    return P;
  };

  vc<int> P;
  if (mode == "full" || mode == "lower") {
    vc<int> Q = calc();
    P.insert(P.end(), all(Q));
  }
  if (mode == "full" || mode == "upper") {
    if (!P.empty()) P.pop_back();
    reverse(all(I));
    vc<int> Q = calc();
    P.insert(P.end(), all(Q));
  }
  if (mode == "upper") reverse(all(P));
  while (len(P) >= 2 && XY[P[0]] == XY[P.back()]) P.pop_back();
  return P;
}
#line 5 "geo/convex_polygon.hpp"

// n=2 は現状サポートしていない
template <typename T>
struct ConvexPolygon {
  using P = Point<T>;
  int n;
  vc<P> point;
  T area2;

  ConvexPolygon(vc<P> point_) : n(len(point_)), point(point_) {
    assert(n >= 3);
    area2 = 0;
    FOR(i, n) {
      int j = nxt_idx(i), k = nxt_idx(j);
      assert((point[j] - point[i]).det(point[k] - point[i]) >= 0);
      area2 += point[i].det(point[j]);
    }
  }

  // 比較関数 comp(i,j)
  template <typename F>
  int periodic_min_comp(F comp) {
    int L = 0, M = n, R = n + n;
    while (1) {
      if (R - L == 2) break;
      int L1 = (L + M) / 2, R1 = (M + R + 1) / 2;
      if (comp(L1 % n, M % n)) { R = M, M = L1; }
      elif (comp(R1 % n, M % n)) { L = M, M = R1; }
      else {
        L = L1, R = R1;
      }
    }
    return M % n;
  }

  int nxt_idx(int i) { return (i + 1 == n ? 0 : i + 1); }
  int prev_idx(int i) { return (i == 0 ? n - 1 : i - 1); }

  // 中：1, 境界：0, 外：-1. test した.
  int side(P p) {
    int L = 1, R = n - 1;
    T a = (point[L] - point[0]).det(p - point[0]);
    T b = (point[R] - point[0]).det(p - point[0]);
    if (a < 0 || b > 0) return -1;
    // p は 0 から見て [L,R] 方向
    while (R - L >= 2) {
      int M = (L + R) / 2;
      T c = (point[M] - point[0]).det(p - point[0]);
      if (c < 0)
        R = M, b = c;
      else
        L = M, a = c;
    }
    T c = (point[R] - point[L]).det(p - point[L]);
    T x = min({a, -b, c});
    if (x < 0) return -1;
    if (x > 0) return 1;
    // on triangle p[0]p[L]p[R]
    if (p == point[0]) return 0;
    if (c != 0 && a == 0 && L != 1) return 1;
    if (c != 0 && b == 0 && R != n - 1) return 1;
    return 0;
  }

  // return {min, idx}. test した.
  pair<T, int> min_dot(P p) {
    int idx = periodic_min_comp([&](int i, int j) -> bool { return point[i].dot(p) < point[j].dot(p); });
    return {point[idx].dot(p), idx};
  }

  // return {max, idx}. test した.
  pair<T, int> max_dot(P p) {
    int idx = periodic_min_comp([&](int i, int j) -> bool { return point[i].dot(p) > point[j].dot(p); });
    return {point[idx].dot(p), idx};
  }

  // p から見える範囲. p 辺に沿って見えるところも見えるとする. test した.
  // 多角形からの反時計順は [l,r] だが p から見た偏角順は [r,l] なので注意
  pair<int, int> visible_range(P p) {
    int a = periodic_min_comp([&](int i, int j) -> bool { return ((point[i] - p).det(point[j] - p) < 0); });
    int b = periodic_min_comp([&](int i, int j) -> bool { return ((point[i] - p).det(point[j] - p) > 0); });
    if ((p - point[a]).det(p - point[prev_idx(a)]) == T(0)) a = prev_idx(a);
    if ((p - point[b]).det(p - point[nxt_idx(b)]) == T(0)) b = nxt_idx(b);
    return {a, b};
  }

  // 線分が「内部と」交わるか
  // https://codeforces.com/contest/1906/problem/D
  bool check_cross(P A, P B) {
    FOR(2) {
      swap(A, B);
      auto [a, b] = visible_range(A);
      if ((point[a] - A).det(B - A) >= 0) return 0;
      if ((point[b] - A).det(B - A) <= 0) return 0;
    }
    return 1;
  }

  vc<T> AREA;

  // point[i,...,j] (inclusive) の面積の 2 倍
  T area_between(int i, int j) {
    assert(i <= j && j <= i + n);
    if (j == i + n) return area2;
    i %= n, j %= n;
    if (i > j) j += n;
    if (AREA.empty()) build_AREA();
    return AREA[j] - AREA[i] + (point[j % n].det(point[i]));
  }

  void build_AREA() {
    AREA.resize(2 * n);
    FOR(i, n) AREA[n + i] = AREA[i] = point[i].det(point[nxt_idx(i)]);
    AREA = cumsum<T>(AREA);
  }

  // 直線の左側の面積. strict に 2 回交わることを仮定.
  // https://codeforces.com/contest/799/problem/G
  T left_area(Line<T> L) {
    static_assert(is_same<T, double>::value || is_same<T, long double>::value);
    Point<T> normal(L.a, L.b);
    int a = min_dot(normal).se;
    int b = max_dot(normal).se;
    if (b < a) b += n;
    assert(L.eval(point[a % n]) < 0 && L.eval(point[b % n]) > 0);
    int p = binary_search([&](int i) -> bool { return L.eval(point[i % n]) < 0; }, a, b);
    int q = binary_search([&](int i) -> bool { return L.eval(point[i % n]) > 0; }, b, a + n);
    T s, t;
    {
      T x = L.eval(point[p % n]);
      T y = L.eval(point[(p + 1) % n]);
      s = x / (x - y);
    }
    {
      T x = L.eval(point[q % n]);
      T y = L.eval(point[(q + 1) % n]);
      t = x / (x - y);
    }
    P A(point[p % n]), B(point[(p + 1) % n]);
    P C(point[q % n]), D(point[(q + 1) % n]);
    P X = B * s + A * (1 - s);
    P Y = D * t + C * (1 - t);
    T ANS = area_between(p, q);
    ANS -= (A - C).det(X - C);
    ANS += (Y - C).det(X - C);
    return ANS;
  }
};

Back to top page