//part of the quick hull algorithm based on http://www.ahristov.com/tutorial/geometry-games/convex-hull.html
int point_location(City A, City B, City P){
int cp1 = (B.x() - A.x())*(P.y()-A.y()) - (B.y()-A.y())*(P.x()-A.x());
if(cp1 > 0)
return 1;
else
return -1;
}
//**in**
//City A the starting point of the line
//City B the ending point of the line
//City point the point to find the distance to
//**out**
//the distance from this line to the point
double point_to_line(Vec2 A, Vec2 B, City point)
{
float diffX = B.x() - A.x();
float diffY = B.y() - A.y();
if ((diffX == 0) && (diffY == 0)){
diffX = point.x() - A.x();
diffY = point.y() - A.y();
return sqrt(diffX * diffX + diffY * diffY);
}
float t = ((point.x() - A.x()) * diffX + (point.y() - A.y()) * diffY) / (diffX * diffX + diffY * diffY);
if (t < 0){
//point is nearest to the first point i.e x1 and y1
diffX = point.x() - A.x();
diffY = point.y() - A.y();
}
else if (t > 1){
//point is nearest to the end point i.e x2 and y2
diffX = point.x() - B.x();
diffY = point.y() - B.y();
}
else
{
//if perpendicular line intersect the line segment.
diffX = point.x() - (A.x() + t * diffX);
diffY = point.y() - (A.y() + t * diffY);
}
//return shortest distance
return sqrt(diffX * diffX + diffY * diffY);
}
//part of the quick hull algorithm based on http://www.ahristov.com/tutorial/geometry-games/convex-hull.html
int distance(City A, City B, City C){
int ABx = B.x() - A.x();
int ABy = B.y() - A.y();
int num = ABx * (A.y() - C.y()) - ABy * (A.x() - C.x());
if (num < 0)
num = -num;
return num;
}