Competitive Programming

/**
 * Author: Ulf Lundstrom
 * Date: 2009-03-21
 * License: CC0
 * Source:
 * Description:\\
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If a unique intersetion point of the lines going through s1,e1 and s2,e2 exists r is set to this point and 1 is returned. If no intersection point exists 0 is returned and if infinitely many exists -1 is returned. If s1==e1 or s2==e2 -1 is returned. The wrong position will be returned if P is Point<int> and the intersection point does not have integer coordinates. Products of three coordinates are used in intermediate steps so watch out for overflow if using int or long long.
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\includegraphics[width=\textwidth]{../content/geometry/lineIntersection}
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 * Status: tested
 * Usage: 
 * 	point<double> intersection;
 * 	if (1 == LineIntersection(s1,e1,s2,e2,intersection))
 * 		cout << "intersection point at " << intersection << endl;
 */
#pragma once

#include "Point.h"

template <class P>
int lineIntersection(const P& s1, const P& e1, const P& s2,
		const P& e2, P& r) {
	if ((e1-s1).cross(e2-s2)) { //if not parallell
		r = s2-(e2-s2)*(e1-s1).cross(s2-s1)/(e1-s1).cross(e2-s2);
		return 1;
	} else
		return -((e1-s1).cross(s2-s1)==0 || s2==e2);
}