检测线的交叉点 [英] Detecting line's intersections
问题描述
我试图来解决这个问题：在屏幕上画线随机然后在所有线路的交叉点创建。我的解决方案：而不是创建实线，我用小点来构造线。每一行是一个称为足尖类的一个实例。我每行实例的所有坐标存储在它自己的数组。为了检测路口，我写了碰撞功能（），以每行的存储坐标进行比较。如果距离＆LT; 10，我改线的厚度10像素打造的气球效应
INT P = 0;
类足尖{
INT X;
诠释Ÿ;
INT speedX;
INT迅速;
浮动的大小= 2;
颜色为c =颜色（随机（255），随机（255），随机（255））;
INT position_stored [] = {};
潘特（INT xPosition位置，诠释yPosition，诠释speed_X，诠释speed_Y）{
X = xPosition位置;
Y = yPosition;
speedX = INT（随机（speed_X））;
SPEEDY = INT（随机（speed_Y））;
} INT B = INT（mouseX）;
INT N = INT（mouseY的）; 无效移动（）{
X = X + speedX;
Y = Y +快速;
如果（X→60）{X = X; Y = Y;}
如果（γ→60）{Y = Y; X = X;}
store_position（）;
} 无效显示（）{
noStroke（）;
填写（C）;
椭圆（X，Y，尺寸，大小）;
}
INT的getX（）{返回X;}
诠释的getY（）{返回Ÿ;}
无效store_position（）{
INT position_stored2 [] =追加（position_stored，的getX（））;
INT position_stored3 [] =追加（position_stored2，的getY（））;
position_stored = position_stored3;
} 无效碰撞（INT [] A）{
的for（int i = 1; I＆LT;则为a.length; I = I + 2）{
INT距离= INT（DIST（X，Y，一个[I-1]中，[I]））;
如果（距离＆LT; 20）{大小= 10;}
其他{大小= 2;}
}
}
}
INT数= 109;
无效设置（）{
大小（600600）;
背景（255）;
的for（int i = 0; I＆LT;数;我++）{
点[I] =新潘特（INT（随机（600）），INT（随机（600）），INT（随机（-6,6）），INT（随机（-6,6）））;
}
}潘特[]积分=新足尖[数字];无效的draw（）{
的for（int i = 0; I＆LT;数;我++）{
对（INT z = 0的; z，其中;号; z ++）{
如果（我== Z）{}
点[I] .collide（分[Z] .position_stored）;
}
点[I]。显示（）;
点[I] .move（）;
}
}
这是一个简单的数学问题。你有什么对每一行都是在飞机上2个点，可以使用这些2点按照此<一个得到直线方程href=\"https://www.khanacademy.org/test-$p$pp/algebra1-brushup/graph-eqns-brushups/v/equation-of-a-line-3\" rel=\"nofollow\">https://www.khanacademy.org/test-$p$pp/algebra1-brushup/graph-eqns-brushups/v/equation-of-a-line-3
那些有2个不同的线方程可以找到他们由G（x）的= F（x）的
问候
I am trying to solve this problem: drawing random lines on the screen then create dots at intersection of all the lines. My solution: instead of creating real lines, I use little dots to construct lines. Each "line" is an instance of a class called Pointe. I store all the coordinates of each line instance in it's own array. In order to detect intersection, I wrote function collide() to compare each line's stored coordinates. If the distance <10, I change the line's thickness to 10 pixel to create the "balloon" effect.
int p = 0;
class Pointe {
int x;
int y;
int speedX;
int speedY;
float size = 2;
color c = color(random(255),random(255),random(255));
int position_stored[] = {};
Pointe(int xPosition, int yPosition, int speed_X, int speed_Y) {
x = xPosition;
y= yPosition;
speedX = int(random(speed_X));
speedY = int(random(speed_Y));
}
int b = int(mouseX);
int n = int(mouseY);
void move() {
x = x + speedX;
y = y + speedY;
if (x>60) {x = x; y = y;}
if (y>60) {y = y; x = x;}
store_position();
}
void display() {
noStroke();
fill(c);
ellipse(x,y,size,size);
}
int getX() {return x;}
int getY() {return y;}
void store_position() {
int position_stored2[] = append(position_stored,getX());
int position_stored3[] = append(position_stored2,getY());
position_stored = position_stored3;
}
void collide(int[] a) {
for (int i = 1; i < a.length ; i=i+2) {
int distance = int(dist(x,y,a[i-1],a[i]));
if (distance < 20) {size = 10;}
else {size = 2;}
}
}
}
int number = 109;
void setup(){
size(600,600);
background(255);
for (int i = 0; i <number; i++) {
points[i] = new Pointe(int(random(600)),int(random(600)),int(random(-6,6)),int(random(-6,6)));
}
}
Pointe[] points = new Pointe[number];
void draw(){
for (int i = 0; i <number; i++) {
for (int z = 0; z <number; z++) {
if (i == z) {}
points[i].collide(points[z].position_stored);
}
points[i].display();
points[i].move();
}
}
This is a simple math problem. What you have for every line are 2 points in the plane, you can get the line equation using those 2 points follow this https://www.khanacademy.org/test-prep/algebra1-brushup/graph-eqns-brushups/v/equation-of-a-line-3 Ones you have the equation of 2 different lines you can find the points where they intersect by G(x) = F(x).
Regards
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