投稿日 11/5/2019最終更新日 11/5/2019

# 問題

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_3_C

点が多角形に内包されるか、されないかまたは多角形の辺上にあるかを判定する問題.

# 解説

詳しいアルゴリズムは以下を参照.

# 解答

C++

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define EPS 0.0000000001
#define fequals(a,b) (fabs((a) - (b)) < EPS)

class Vector2 {
public:
  double x, y;
  
  Vector2(double x = 0, double y = 0): x(x), y(y) {}
  
  Vector2 operator + (const Vector2 v) const { return Vector2(x + v.x, y + v.y); }
  Vector2 operator - (const Vector2 v) const { return Vector2(x - v.x, y - v.y); }
  Vector2 operator * (const double k) const { return Vector2(x * k, y * k); }
  Vector2 operator / (const double k) const { return Vector2(x / k, y / k); }
  
  double length() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  double dot (Vector2 const v) { return x * v.x + y * v.y; }
  double cross (Vector2 const v) { return x * v.y - y * v.x; }
  
  bool parallel(Vector2 &other) {
    return fequals(0, cross(other));
  }
  
  bool orthogonal(Vector2 &other) {
    return fequals(0, dot(other));
  }
  
  bool operator < (const Vector2 &v) {
    return x != v.x ? x < v.x : y < v.y;
  }
  
  bool operator == (const Vector2 &v) {
    return fabs(x - v.x) < EPS && fabs(y - v.y) < EPS;
  }
};

ostream & operator << (ostream & out, Vector2 const & v) { 
  out<< "Vector2(" << v.x << ", " << v.y << ')';
  return out;
}

istream & operator >> (istream & in, Vector2 & v) { 
  double x, y;
  in >> x;
  in >> y;
  v.x = x;
  v.y = y;
  return in;
}

#define MAX_N 101
#define MAX_Q 1001

Int N, Q;
Vector2 polygon[MAX_N];
Vector2 point;

Int contains(Vector2 &point) {
  bool in = false;
  loop(n,0,N) {
    Vector2 a = polygon[n] - point;
    Vector2 b = polygon[(n+1) % N] - point;
    if (fabs(a.cross(b)) < EPS && a.dot(b) < EPS) return 1;
    if (a.y > b.y) swap(a, b);
    if (a.y < EPS && b.y > EPS && a.cross(b) > EPS) in = !in;
  }
  
  return in ? 2 : 0;
}

void input() {
  cin >> N;
  loop(n,0,N) cin >> polygon[n];
  cin >> Q;
  while (cin >> point) {
    cout << contains(point) << endl;
  }
}

int main() {
  cout.precision(15);
  input();
}