完善盒子相交测试和线段相交测试的代码文档.

分离`DoIntersect`代码

src/Editor/SelectLine.ts

@@ -2,7 +2,7 @@ import { Frustum, Matrix4, Object3D, OrthographicCamera, PerspectiveCamera, Vect
 import { PreViewer } from '../GraphicsSystem/PreViewer';
 import { Viewer } from '../GraphicsSystem/Viewer';
 import { SelectSetBase } from './SelectBase';
-import { doIntersect } from '../Geometry/CheckIntersect';
+import { doIntersect } from "../Geometry/DoIntersect";
 
 export class SelectLine extends SelectSetBase
 {

src/Geometry/CheckIntersect.ts

@@ -1,6 +1,15 @@
 import { Box2, Vector2, Math as Math3 } from "three";
+import { doIntersect } from "./DoIntersect";
+
+/*
+功能:盒子相交检测,快速判断正矩形(未旋转的)与直线和圆是否有交点.
+1.使用裁剪算法优化判断速度.
+参考:https://zh.wikipedia.org/wiki/%E7%A7%91%E6%81%A9%EF%BC%8D%E8%8B%8F%E6%B3%BD%E5%85%B0%E7%AE%97%E6%B3%95
+2.使用快速判断直线是否有交点提高速度.
+参考:doIntersect方法.
+*/
 
-interface Vec2
+export interface Vec2
 {
     x: number;
     y: number;
@@ -11,23 +20,26 @@ const LEFT = 1, RIGHT = 2, DOWN = 4, TOP = 8;
 // bounded diagonally by (xmin, ymin), and (xmax, ymax)
 
 // ASSUME THAT xmax, xmin, ymax and ymin are global constants.
-
 function ComputeOutCode(x: number, y: number, box: Box2): number
 {
     let code = 0;
     if (x < box.min.x)           // to the left of clip window
-        code |= 1;
+        code |= LEFT;
     else if (x > box.max.x)      // to the right of clip window
-        code |= 2;
+        code |= RIGHT;
 
     if (y < box.min.y)           // below the clip window
-        code |= 4;
+        code |= DOWN;
     else if (y > box.max.y)      // above the clip window
-        code |= 8;
+        code |= TOP;
 
     return code;
 }
 
+
+/**
+ * 盒子相交测试,快速判断盒子和直线或者圆是否有相交
+ */
 export class BoxCheckIntersect
 {
     p1: Vec2;
@@ -75,66 +87,3 @@ export class BoxCheckIntersect
         return ((nearestX - cen.x) ** 2 + (nearestY - cen.y) ** 2) <= radius ** 2;
     }
 }
-
-
-//判断线段是否存在交点 ref: https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
-
-// Given three colinear points p, q, r, the function checks if
-// point q lies on line segment 'pr'
-function onSegment(p: Vec2, q: Vec2, r: Vec2): boolean
-{
-    if (q.x <= Math.max(p.x, r.x) && q.x >= Math.min(p.x, r.x) &&
-        q.y <= Math.max(p.y, r.y) && q.y >= Math.min(p.y, r.y))
-        return true;
-
-    return false;
-}
-
-// To find orientation of ordered triplet (p, q, r).
-// The function returns following values
-// 0 --> p, q and r are colinear
-// 1 --> Clockwise
-// 2 --> Counterclockwise
-function orientation(p: Vec2, q: Vec2, r: Vec2): number
-{
-    // See https://www.geeksforgeeks.org/orientation-3-ordered-points/
-    // for details of below formula.
-    let val = (q.y - p.y) * (r.x - q.x) -
-        (q.x - p.x) * (r.y - q.y);
-
-    if (val == 0) return 0;  // colinear
-
-    return (val > 0) ? 1 : 2; // clock or counterclock wise
-}
-
-/**
- * 判断`p1q1`和`p2q2`是否相交.
- */
-export function doIntersect(p1: Vec2, q1: Vec2, p2: Vec2, q2: Vec2): boolean
-{
-    // Find the four orientations needed for general and
-    // special cases
-    let o1 = orientation(p1, q1, p2);
-    let o2 = orientation(p1, q1, q2);
-    let o3 = orientation(p2, q2, p1);
-    let o4 = orientation(p2, q2, q1);
-
-    // General case
-    if (o1 != o2 && o3 != o4)
-        return true;
-
-    // Special Cases
-    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
-    if (o1 == 0 && onSegment(p1, p2, q1)) return true;
-
-    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
-    if (o2 == 0 && onSegment(p1, q2, q1)) return true;
-
-    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
-    if (o3 == 0 && onSegment(p2, p1, q2)) return true;
-
-    // p2, q2 and q1 are colinear and q1 lies on segment p2q2
-    if (o4 == 0 && onSegment(p2, q1, q2)) return true;
-
-    return false; // Doesn't fall in any of the above cases
-}

src/Geometry/DoIntersect.ts

@@ -0,0 +1,59 @@
+import { Vec2 } from "./CheckIntersect";
+/*
+功能:判断线段是否存在交点
+ref:https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/
+*/
+
+// Given three colinear points p, q, r, the function checks if
+// point q lies on line segment 'pr'
+function onSegment(p: Vec2, q: Vec2, r: Vec2): boolean
+{
+    if (q.x <= Math.max(p.x, r.x) && q.x >= Math.min(p.x, r.x) &&
+        q.y <= Math.max(p.y, r.y) && q.y >= Math.min(p.y, r.y))
+        return true;
+    return false;
+}
+// To find orientation of ordered triplet (p, q, r).
+// The function returns following values
+// 0 --> p, q and r are colinear
+// 1 --> Clockwise
+// 2 --> Counterclockwise
+function orientation(p: Vec2, q: Vec2, r: Vec2): number
+{
+    // See https://www.geeksforgeeks.org/orientation-3-ordered-points/
+    // for details of below formula.
+    let val = (q.y - p.y) * (r.x - q.x) -
+        (q.x - p.x) * (r.y - q.y);
+    if (val == 0)
+        return 0; // colinear
+    return (val > 0) ? 1 : 2; // clock or counterclock wise
+}
+/**
+ * 判断线段`p1q1`和线段`p2q2`是否相交.
+ */
+export function doIntersect(p1: Vec2, q1: Vec2, p2: Vec2, q2: Vec2): boolean
+{
+    // Find the four orientations needed for general and
+    // special cases
+    let o1 = orientation(p1, q1, p2);
+    let o2 = orientation(p1, q1, q2);
+    let o3 = orientation(p2, q2, p1);
+    let o4 = orientation(p2, q2, q1);
+    // General case
+    if (o1 !== o2 && o3 !== o4)
+        return true;
+    // Special Cases
+    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
+    if (o1 === 0 && onSegment(p1, p2, q1))
+        return true;
+    // p1, q1 and q2 are colinear and q2 lies on segment p1q1
+    if (o2 === 0 && onSegment(p1, q2, q1))
+        return true;
+    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
+    if (o3 === 0 && onSegment(p2, p1, q2))
+        return true;
+    // p2, q2 and q1 are colinear and q1 lies on segment p2q2
+    if (o4 === 0 && onSegment(p2, q1, q2))
+        return true;
+    return false; // Doesn't fall in any of the above cases
+}