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@ -57,14 +57,17 @@ export class BoxCheckIntersect
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this.box = box;
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}
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//直线与盒子相交,或者被盒子包含.
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//直线与盒子相交,或者被盒子包含. Cohen–Sutherland裁剪算法
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IsIntersectLine(p1: Vec2, p2: Vec2): boolean
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{
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let code1 = ComputeOutCode(p1.x, p1.y, this.box);
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let code2 = ComputeOutCode(p2.x, p2.y, this.box);
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//按位AND不为0:两个点共享一个外部区域(LEFT,RIGHT,TOP或BOTTOM),因此两个点都必须在窗口外部
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if (code1 & code2) return false;
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let code = code1 | code2;
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if (code1 === 0 || code2 === 0 || code === 3 || code === 12)
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if (code1 === 0 || code2 === 0 || code === 3 || code === 12)//点1点2在矩形内,或者竖直贯穿,水平贯穿
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return true;
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if ((code & TOP) && doIntersect(p1, p2, this.p3, this.p4))
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@ -87,3 +90,89 @@ export class BoxCheckIntersect
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return ((nearestX - cen.x) ** 2 + (nearestY - cen.y) ** 2) <= radius ** 2;
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}
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}
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//https://en.wikipedia.org/wiki/Cohen%E2%80%93Sutherland_algorithm
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function CohenSutherlandLineClip(box: Box2, p1: Vec2, p2: Vec2)
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{
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let [x0, y0] = [p1.x, p1.y];
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let [x1, y1] = [p2.x, p2.y];
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let xmin = box.min.x;
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let ymin = box.min.y;
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let xmax = box.max.x;
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let ymax = box.max.y;
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// compute outcodes for P0, P1, and whatever point lies outside the clip rectangle
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let outcode0 = ComputeOutCode(x0, y0, box);
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let outcode1 = ComputeOutCode(x1, y1, box);
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let accept = false;
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while (true)
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{
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if (!(outcode0 | outcode1))
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{
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// bitwise OR is 0: both points inside window; trivially accept and exit loop
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accept = true;
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break;
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}
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else if (outcode0 & outcode1)
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{
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// bitwise AND is not 0: both points share an outside zone (LEFT, RIGHT, TOP,
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// or BOTTOM), so both must be outside window; exit loop (accept is false)
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break;
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}
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else
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{
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// failed both tests, so calculate the line segment to clip
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// from an outside point to an intersection with clip edge
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let x: number, y: number;
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// At least one endpoint is outside the clip rectangle; pick it.
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let outcodeOut = outcode1 > outcode0 ? outcode1 : outcode0;
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// Now find the intersection point;
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// use formulas:
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// slope = (y1 - y0) / (x1 - x0)
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// x = x0 + (1 / slope) * (ym - y0), where ym is ymin or ymax
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// y = y0 + slope * (xm - x0), where xm is xmin or xmax
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// No need to worry about divide-by-zero because, in each case, the
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// outcode bit being tested guarantees the denominator is non-zero
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if (outcodeOut & TOP)
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{ // point is above the clip window
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x = x0 + (x1 - x0) * (ymax - y0) / (y1 - y0);
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y = ymax;
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}
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else if (outcodeOut & DOWN)
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{ // point is below the clip window
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x = x0 + (x1 - x0) * (ymin - y0) / (y1 - y0);
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y = ymin;
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}
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else if (outcodeOut & RIGHT)
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{ // point is to the right of clip window
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y = y0 + (y1 - y0) * (xmax - x0) / (x1 - x0);
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x = xmax;
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}
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else if (outcodeOut & LEFT)
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{ // point is to the left of clip window
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y = y0 + (y1 - y0) * (xmin - x0) / (x1 - x0);
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x = xmin;
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}
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// Now we move outside point to intersection point to clip
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// and get ready for next pass.
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if (outcodeOut === outcode0)
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{
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x0 = x;
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y0 = y;
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outcode0 = ComputeOutCode(x0, y0, box);
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}
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else
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{
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x1 = x;
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y1 = y;
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outcode1 = ComputeOutCode(x1, y1, box);
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}
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}
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}
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}
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ChenX
3 years ago