我为我的平台游戏编写了地图编辑器。我正在使用 SFML。地图由多边形组成 - ConvexShapes。我需要通过单击它们来添加 ConvexShapes 的选择。我知道我可以使用cv.getLocalBounds() 来获取矩形并接下来检查，但我需要更精确的解决方案。如何检查点击的点是否属于任何形状？
【问题讨论】：
基于问题 cmets link，这就是我得到的。
使用那里描述的算法，我们可以确定给定点是否在一个形状内，计算每边有多少交叉点
如果它们的数量每边都是奇数，则该点属于形状。那么简单：
bool contains(sf::ConvexShape shape, sf::Vector2f point){
std::vector<sf::Vector2f> intersectPoints = getIntersectionPoints(shape, point);
int nodesAtLeft = 0;
int nodesAtRight = 0;
for (sf::Vector2f po : intersectPoints){
if (po.x < point.x){
nodesAtLeft++;
}
else if(po.x > point.x){
nodesAtRight++;
}
}
return ((nodesAtLeft % 2) == 1) && ((nodesAtRight % 2) == 1);
}
那么，我们如何获得这些交点？我们应该检查形状的每一侧，其中 交叉点 与由我们给定的点确定的水平线。请注意，交叉点可能远离形状，因为交叉点的计算考虑整条线，不是线段
所以，一旦我们得到了那个交叉点，我们应该检查它是否属于这个段。
std::vector<sf::Vector2f> getIntersectionPoints(sf::ConvexShape shape, sf::Vector2f point){
std::vector<sf::Vector2f> intersectPoints;
sf::Vector2f p;
bool crossingLine; // This will be used to avoid duplicated points on special cases
if (shape.getPointCount() < 3){
return intersectPoints;
}
sf::FloatRect bounds = shape.getLocalBounds();
// To determine horizontal line, we use two points, one at leftmost side of the shape (in fact, its bound) and the other at rightmost side
Line pointLine, shapeLine;
pointLine.p1 = sf::Vector2f(bounds.left, point.y);
pointLine.p2 = sf::Vector2f(bounds.left + bounds.width, point.y);
unsigned int nPoints = shape.getPointCount();
for (int i = 0; i < nPoints; ++i){
try{
shapeLine.p1 = shape.getPoint(i % nPoints); // Last one will be nPoints-1
shapeLine.p2 = shape.getPoint((i + 1) % nPoints); // So this one must be 0 in order to check last side (returning to origin)
crossingLine = (shapeLine.p1.y >= point.y && shapeLine.p2.y <= point.y) || (shapeLine.p2.y >= point.y && shapeLine.p1.y <= point.y);
p = intersection(shapeLine, pointLine);
if (crossingLine && shapeLine.contains(p))
intersectPoints.push_back(p);
}
catch (std::runtime_error e){
}
}
return intersectPoints;
}
我认为检查点 C 是否属于段（由点 A 和 B 定义）的最简单方法是通过执行距离检查，例如：
distance(A,C) + distance(C,B) == distance(A,B)
但是由于这个可能会过于严格，我已经对其进行了调整以考虑一点误差范围：
abs((distance(A, C) + distance(C, B)) - distance(A, B)) < margin
在我忘记之前，Line 是这样定义的
struct Line{
sf::Vector2f p1;
sf::Vector2f p2;
bool contains(sf::Vector2f point) const{
float margin = 0.1;
return std::abs((distance(p1, point) + distance(point, p2)) - distance(p1, p2)) < margin;
}
};
有了这个，现在唯一剩下的就是计算两条给定线之间的交点。真诚的，这个我就不解释了（主要是我刚刚从wikipedia复制过来的）
sf::Vector2f intersection(Line lineA, Line lineB){
int x1 = lineA.p1.x;
int y1 = lineA.p1.y;
int x2 = lineA.p2.x;
int y2 = lineA.p2.y;
int x3 = lineB.p1.x;
int y3 = lineB.p1.y;
int x4 = lineB.p2.x;
int y4 = lineB.p2.y;
try{
double retX = ((x1*y2 - y1*x2)*(x3 - x4) - (x1 - x2)*(x3*y4 - y3*x4)) / ((x1 - x2)*(y3 - y4) - (y1 - y2)*(x3 - x4));
double retY = ((x1*y2 - y1*x2)*(y3 - y4) - (y1 - y2)*(x3*y4 - y3*x4)) / ((x1 - x2)*(y3 - y4) - (y1 - y2)*(x3 - x4));
return sf::Vector2f(retX, retY);
}
catch (std::exception){
throw new std::exception("");
}
}
如果两条线平行或同一条线，两个分母都为零，会抛出DivideByZero异常，没有交集就不是问题。
我还做了一个 sn-p 来测试这个：
int main()
{
sf::RenderWindow v(sf::VideoMode(600,400), "SFML");
sf::ConvexShape shape;
std::vector<sf::Vector2i> points;
std::vector<sf::CircleShape> intPoints;
shape.setPointCount(0);
shape.setOutlineColor(sf::Color::Blue);
shape.setFillColor(sf::Color::Black);
shape.setOutlineThickness(1);
while (v.isOpen()){
sf::Event event;
while (v.pollEvent(event)){
if (event.type == sf::Event::Closed)
v.close();
else if (event.type == sf::Event::MouseButtonPressed){
if (event.mouseButton.button == sf::Mouse::Button::Left){
// Add a point to the shape
intPoints.clear();
sf::Vector2i p = sf::Mouse::getPosition(v);
points.push_back(p);
shape.setPointCount(points.size());
for (int i = 0; i < points.size(); ++i){
shape.setPoint(i, sf::Vector2f(points[i]));
}
}
else if (event.mouseButton.button == sf::Mouse::Button::Right){
// Delete shape
points.clear();
intPoints.clear();
shape.setPointCount(0);
}
else if (event.mouseButton.button == sf::Mouse::Button::Middle){
// Set testing point
intPoints.clear();
sf::Vector2i p = sf::Mouse::getPosition(v);
if (contains(shape, sf::Vector2f(p))){
std::cout << "Point inside shape" << std::endl;
}
else{
std::cout << "Point outside shape" << std::endl;
}
auto v = getIntersectionPoints(shape, sf::Vector2f(p));
for (sf::Vector2f po : v){
sf::CircleShape c(2);
c.setFillColor(sf::Color::Green);
c.setOrigin(1, 1);
c.setPosition(po);
intPoints.push_back(c);
}
// testing point added too, to be visualized
sf::CircleShape c(2);
c.setFillColor(sf::Color::Red);
c.setOrigin(1, 1);
c.setPosition(sf::Vector2f(p));
intPoints.push_back(c);
}
}
}
v.clear();
v.draw(shape);
for (sf::CircleShape c : intPoints){
v.draw(c);
}
v.display();
}
return 0;
}
一些捕获：
可能是一篇很长的帖子，但我已经尽力说清楚了。
【讨论】：