我一直在看Emanuele Feronato的字符串避免代码(下面的代码和链接),并尝试调整它,以便当字符串与自身相交时 – 形成一个闭环 – 它只填充该区域
see image.我的数学很可怕而且我在挣扎理解三角形.有什么建议?谢谢.
这是一个帮助澄清的图表 – http://samtripp.com/eg.png
package {
import flash.display.Sprite;
import flash.events.Event;
import flash.geom.Point;
public class Main extends Sprite {
private var tailLenght:Number=50;
private var tailNodes:Number=10;
private var head:headMc=new headMc();
private var tailCanvas:Sprite=new Sprite();
private var nodes:Vector.<Point>=new Vector.<Point>();
public function Main() {
addChild(head);
head.x=320;
head.y=240;
addChild(tailCanvas);
for (var i:int=0; i<tailNodes; i++) {
nodes[i]=new Point(head.x,head.y);
}
addEventListener(Event.ENTER_FRAME,update);
}
private function update(e:Event):void {
head.x=mouseX;
head.y=mouseY;
tailCanvas.graphics.clear();
tailCanvas.graphics.lineStyle(2,0x00ff00);
tailCanvas.graphics.moveTo(head.x,head.y);
nodes[0]=new Point(head.x,head.y);
for (var i:int=1; i<tailNodes-1; i++) {
var nodeAngle:Number=Math.atan2(nodes[i].y-nodes[i-1].y,nodes[i].x-nodes[i-1].x);
nodes[i]=new Point(nodes[i-1].x+tailLenght*Math.cos(nodeAngle),nodes[i-1].y+tailLenght*Math.sin(nodeAngle));
if (i<tailNodes-2) {
for (var j:int=i-1; j>=1; j--) {
var p:Point=lineIntersection(nodes[j],nodes[j-1],nodes[i],nodes[i+1]);
if (p!=null) {
tailCanvas.graphics.beginFill(0x000000);
tailCanvas.graphics.drawCircle(p.x,p.y,4);
tailCanvas.graphics.endFill();
tailCanvas.graphics.moveTo(nodes[i-1].x,nodes[i-1].y);
}
}
}
tailCanvas.graphics.lineTo(nodes[i].x,nodes[i].y);
}
}
private function lineIntersection(p1:Point,p2:Point,p3:Point,p4:Point):Point {
var x1:Number=p1.x;
var x2:Number=p2.x;
var x3:Number=p3.x;
var x4:Number=p4.x;
var y1:Number=p1.y;
var y2:Number=p2.y;
var y3:Number=p3.y;
var y4:Number=p4.y;
var px:Number=((x1*y2-y1*x2)*(x3-x4)-(x1-x2)*(x3*y4-y3*x4))/((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4));
var py:Number=((x1*y2-y1*x2)*(y3-y4)-(y1-y2)*(x3*y4-y3*x4))/((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4));
var segment1Len:Number=Math.pow(p1.x-p2.x,2)+Math.pow(p1.y-p2.y,2);
var segment2Len:Number=Math.pow(p3.x-p4.x,2)+Math.pow(p3.y-p4.y,2);
if (Math.pow(p1.x-px,2)+Math.pow(p1.y-py,2)>segment1Len) {
return null;
}
if (Math.pow(p2.x-px,2)+Math.pow(p2.y-py,2)>segment1Len) {
return null;
}
if (Math.pow(p3.x-px,2)+Math.pow(p3.y-py,2)>segment2Len) {
return null;
}
if (Math.pow(p4.x-px,2)+Math.pow(p4.y-py,2)>segment2Len) {
return null;
}
return new Point(px,py);
}
}
}
最佳答案 在该网站上有很多很好的教程,我推荐它.
刚刚编写了一个快速功能,可以满足您的需求:
private function fillIntersection(p,startN,endN):void{
tailCanvas.graphics.beginFill(0x0000FF,0.5);
tailCanvas.graphics.moveTo(p.x,p.y);
for (var i:int=endN; i<startN; i++) {
tailCanvas.graphics.lineTo(nodes[i].x,nodes[i].y);
}
tailCanvas.graphics.lineTo(nodes[startN].x,nodes[startN].y);
tailCanvas.graphics.endFill();
}
你应该在以下之后运行:if(p!= null){(第34行)
像这样:
if (p!=null) {
fillIntersection(p,i,j);
tailCanvas.graphics.beginFill(0x000000);
为了解释它是如何工作的,函数给出了要填充的线的点,以及碰撞的确切点,然后它遍历它们并绘制一个填充的多边形.
(我确定有更好的方法,但它的工作原理很容易理解)