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geometry.cpp
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geometry.cpp
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#include <iostream>
#include <vector>
#include <cmath>
#include <cassert>
using namespace std;
double INF = 1e100;
double EPS = 1e-12;
struct PT {
double x, y;
PT() {}
PT(double x, double y) : x(x), y(y) {}
PT(const PT &p) : x(p.x), y(p.y) {}
PT operator + (const PT &p) const { return PT(x+p.x, y+p.y); }
PT operator - (const PT &p) const { return PT(x-p.x, y-p.y); }
PT operator * (double c) const { return PT(x*c, y*c ); }
PT operator / (double c) const { return PT(x/c, y/c ); }
};
double dot(PT p, PT q) { return p.x*q.x+p.y*q.y; }
double dist2(PT p, PT q) { return dot(p-q,p-q); }
double cross(PT p, PT q) { return p.x*q.y-p.y*q.x; }
ostream &operator<<(ostream &os, const PT &p) {
return os << "(" << p.x << "," << p.y << ")";
}
// if movement from a to b to c is done in a CW path returns 1
// else if it's CCW returns -1 and if they make a line returns 0
int IsCWTurn(PT a, PT b, PT c) {
double r = cross((b - c), (a - c));
return (fabs(r) < EPS)? 0: (r > 0)? 1: -1;
}
// rotate a point CCW or CW around the origin
PT RotateCCW90(PT p) { return PT(-p.y,p.x); }
PT RotateCW90(PT p) { return PT(p.y,-p.x); }
PT RotateCCW(PT p, double t) {
return PT(p.x*cos(t)-p.y*sin(t), p.x*sin(t)+p.y*cos(t));
}
// project point c onto line through a and b
// assuming a != b
PT ProjectPointLine(PT a, PT b, PT c) {
return a + (b-a)*dot(c-a, b-a)/dot(b-a, b-a);
}
// project point c onto line segment through a and b
PT ProjectPointSegment(PT a, PT b, PT c) {
double r = dot(b-a,b-a);
if (fabs(r) < EPS) return a;
r = dot(c-a, b-a)/r;
return (r < 0)? a: (r > 1)? b: a + (b - a)*r;
}
// compute distance from c to segment between a and b
double DistancePointSegment(PT a, PT b, PT c) {
return sqrt(dist2(c, ProjectPointSegment(a, b, c)));
}
// compute distance between point (x,y,z) and plane ax+by+cz=d
double DistancePointPlane(double x, double y, double z, double a, double b,
double c, double d) {
return fabs(a*x+b*y+c*z-d)/sqrt(a*a+b*b+c*c);
}
// determine if lines from a to b and c to d are parallel or collinear
bool LinesParallel(PT a, PT b, PT c, PT d) {
return fabs(cross(b-a, c-d)) < EPS;
}
bool LinesCollinear(PT a, PT b, PT c, PT d) {
return LinesParallel(a, b, c, d)
&& fabs(cross(a-b, a-c)) < EPS
&& fabs(cross(c-d, c-a)) < EPS;
}
// determine if line segment from a to b intersects with
// line segment from c to d
bool SegmentsIntersect(PT a, PT b, PT c, PT d) {
if (LinesCollinear(a, b, c, d)) {
if (dist2(a, c) < EPS || dist2(a, d) < EPS ||
dist2(b, c) < EPS || dist2(b, d) < EPS) return true;
if (dot(c-a, c-b) > 0 && dot(d-a, d-b) > 0 && dot(c-b, d-b) > 0)
return false;
return true;
}
if (cross(d-a, b-a) * cross(c-a, b-a) > 0) return false;
if (cross(a-c, d-c) * cross(b-c, d-c) > 0) return false;
return true;
}
// compute intersection of line passing through a and b
// with line passing through c and d, assuming that unique
// intersection exists; for segment intersection, check if
// segments intersect first
PT ComputeLineIntersection(PT a, PT b, PT c, PT d) {
b=b-a; d=c-d; c=c-a;
assert(dot(b, b) > EPS && dot(d, d) > EPS);
return a + b*cross(c, d)/cross(b, d);
}
// compute center of circle given three points
PT ComputeCircleCenter(PT a, PT b, PT c) {
b=(a+b)/2;
c=(a+c)/2;
return ComputeLineIntersection(b, b+RotateCW90(a-b), c, c+RotateCW90(a-c));
}
// determine if point is in a possibly non-convex polygon (by William
// Randolph Franklin); returns 1 for strictly interior points, 0 for
// strictly exterior points, and 0 or 1 for the remaining points.
// Note that it is possible to convert this into an *exact* test using
// integer arithmetic by taking care of the division appropriately
// (making sure to deal with signs properly) and then by writing exact
// tests for checking point on polygon boundary
bool PointInPolygon(const vector<PT> &p, PT q) {
bool c = 0;
for (int i = 0; i < p.size(); i++) {
int j = (i+1)%p.size();
if (((p[i].y <= q.y && q.y < p[j].y) || (p[j].y <= q.y && q.y < p[i].y)) &&
q.x < p[i].x + (p[j].x - p[i].x) * (q.y - p[i].y) / (p[j].y - p[i].y))
c = !c;
}
return c;
}
// determine if point is on the boundary of a polygon
bool PointOnPolygon(const vector<PT> &p, PT q) {
for (int i = 0; i < p.size(); i++)
if (dist2(ProjectPointSegment(p[i], p[(i+1)%p.size()], q), q) < EPS)
return true;
return false;
}
// compute intersection of line through points a and b with
// circle centered at c with radius r > 0
vector<PT> CircleLineIntersection(PT a, PT b, PT c, double r) {
vector<PT> ret;
b = b-a;
a = a-c;
double A = dot(b, b);
double B = dot(a, b);
double C = dot(a, a) - r*r;
double D = B*B - A*C;
if (D < -EPS) return ret;
ret.push_back(c+a+b*(-B+sqrt(D+EPS))/A);
if (D > EPS)
ret.push_back(c+a+b*(-B-sqrt(D))/A);
return ret;
}
// compute intersection of circle centered at a with radius r
// with circle centered at b with radius R
vector<PT> CircleCircleIntersection(PT a, PT b, double r, double R) {
vector<PT> ret;
double d = sqrt(dist2(a, b));
if (d > r+R || d+min(r, R) < max(r, R)) return ret;
double x = (d*d-R*R+r*r)/(2*d);
double y = sqrt(r*r-x*x);
PT v = (b-a)/d;
ret.push_back(a+v*x + RotateCCW90(v)*y);
if (y > 0)
ret.push_back(a+v*x - RotateCCW90(v)*y);
return ret;
}
// This code computes the area or centroid of a (possibly nonconvex)
// polygon, assuming that the coordinates are listed in a clockwise or
// counterclockwise fashion. Note that the centroid is often known as
// the "center of gravity" or "center of mass".
double ComputeSignedArea(const vector<PT> &p) {
double area = 0;
for(int i = 0; i < p.size(); i++) {
int j = (i+1) % p.size();
area += p[i].x*p[j].y - p[j].x*p[i].y;
}
return area / 2.0;
}
double ComputeArea(const vector<PT> &p) {
return fabs(ComputeSignedArea(p));
}
PT ComputeCentroid(const vector<PT> &p) {
PT c(0,0);
double scale = 6.0 * ComputeSignedArea(p);
for (int i = 0; i < p.size(); i++){
int j = (i+1) % p.size();
c = c + (p[i]+p[j])*(p[i].x*p[j].y - p[j].x*p[i].y);
}
return c / scale;
}
// tests whether or not a given polygon (in CW or CCW order) is simple
bool IsSimple(const vector<PT> &p) {
for (int i = 0; i < p.size(); i++) {
for (int k = i+1; k < p.size(); k++) {
int j = (i+1) % p.size();
int l = (k+1) % p.size();
if (i == l || j == k) continue;
if (SegmentsIntersect(p[i], p[j], p[k], p[l]))
return false;
}
}
return true;
}