## Algoritmo de lineIntersection.

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/**
* Author: Ulf Lundstrom
* Date: 2009-03-21
* License: CC0
* Source:
* Description:\\
\begin{minipage}{75mm}
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.
\end{minipage}
\begin{minipage}{15mm}
\includegraphics[width=\textwidth]{../content/geometry/lineIntersection}
\end{minipage}
* 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);
}


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