我如何确定两条直线是否相交,如果相交,在x,y点处?
当前回答
如果矩形的每条边都是一条线段,并且用户绘制的部分也是一条线段,那么您只需检查用户绘制的线段是否与四条边线段相交。这应该是一个相当简单的练习,给定每个段的起点和终点。
其他回答
只是想提一下,一个很好的解释和明确的解决方案可以在数字食谱系列中找到。我有这本书的第三版,答案在1117页21.4节。另一种不同命名的解决方案可以在玛丽娜·加夫里洛娃(Marina Gavrilova)的论文中找到。在我看来,她的解决办法要简单一些。
我的实现如下:
bool NuGeometry::IsBetween(const double& x0, const double& x, const double& x1){
return (x >= x0) && (x <= x1);
}
bool NuGeometry::FindIntersection(const double& x0, const double& y0,
const double& x1, const double& y1,
const double& a0, const double& b0,
const double& a1, const double& b1,
double& xy, double& ab) {
// four endpoints are x0, y0 & x1,y1 & a0,b0 & a1,b1
// returned values xy and ab are the fractional distance along xy and ab
// and are only defined when the result is true
bool partial = false;
double denom = (b0 - b1) * (x0 - x1) - (y0 - y1) * (a0 - a1);
if (denom == 0) {
xy = -1;
ab = -1;
} else {
xy = (a0 * (y1 - b1) + a1 * (b0 - y1) + x1 * (b1 - b0)) / denom;
partial = NuGeometry::IsBetween(0, xy, 1);
if (partial) {
// no point calculating this unless xy is between 0 & 1
ab = (y1 * (x0 - a1) + b1 * (x1 - x0) + y0 * (a1 - x1)) / denom;
}
}
if ( partial && NuGeometry::IsBetween(0, ab, 1)) {
ab = 1-ab;
xy = 1-xy;
return true;
} else return false;
}
我将Kris的答案移植到JavaScript。在尝试了许多不同的答案后,他给出了正确的观点。我以为我要疯了,因为我没有得到我需要的分数。
function getLineLineCollision(p0, p1, p2, p3) {
var s1, s2;
s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
s2 = {x: p3.x - p2.x, y: p3.y - p2.y};
var s10_x = p1.x - p0.x;
var s10_y = p1.y - p0.y;
var s32_x = p3.x - p2.x;
var s32_y = p3.y - p2.y;
var denom = s10_x * s32_y - s32_x * s10_y;
if(denom == 0) {
return false;
}
var denom_positive = denom > 0;
var s02_x = p0.x - p2.x;
var s02_y = p0.y - p2.y;
var s_numer = s10_x * s02_y - s10_y * s02_x;
if((s_numer < 0) == denom_positive) {
return false;
}
var t_numer = s32_x * s02_y - s32_y * s02_x;
if((t_numer < 0) == denom_positive) {
return false;
}
if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
return false;
}
var t = t_numer / denom;
var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
return p;
}
根据t3chb0t的答案:
int intersezione_linee(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
//L1: estremi (x1,y1)(x2,y2) L2: estremi (x3,y3)(x3,y3)
int d;
d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4);
if(!d)
return 0;
p_x = ((x1*y2-y1*x2)*(x3-x4) - (x1-x2)*(x3*y4-y3*x4))/d;
p_y = ((x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4))/d;
return 1;
}
int in_bounding_box(int x1, int y1, int x2, int y2, int p_x, int p_y)
{
return p_x>=x1 && p_x<=x2 && p_y>=y1 && p_y<=y2;
}
int intersezione_segmenti(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
if (!intersezione_linee(x1,y1,x2,y2,x3,y3,x4,y4,p_x,p_y))
return 0;
return in_bounding_box(x1,y1,x2,y2,p_x,p_y) && in_bounding_box(x3,y3,x4,y4,p_x,p_y);
}
iMalc回答的Python版本:
def find_intersection( p0, p1, p2, p3 ) :
s10_x = p1[0] - p0[0]
s10_y = p1[1] - p0[1]
s32_x = p3[0] - p2[0]
s32_y = p3[1] - p2[1]
denom = s10_x * s32_y - s32_x * s10_y
if denom == 0 : return None # collinear
denom_is_positive = denom > 0
s02_x = p0[0] - p2[0]
s02_y = p0[1] - p2[1]
s_numer = s10_x * s02_y - s10_y * s02_x
if (s_numer < 0) == denom_is_positive : return None # no collision
t_numer = s32_x * s02_y - s32_y * s02_x
if (t_numer < 0) == denom_is_positive : return None # no collision
if (s_numer > denom) == denom_is_positive or (t_numer > denom) == denom_is_positive : return None # no collision
# collision detected
t = t_numer / denom
intersection_point = [ p0[0] + (t * s10_x), p0[1] + (t * s10_y) ]
return intersection_point
这对我来说很有效。从这里拍的。
// calculates intersection and checks for parallel lines.
// also checks that the intersection point is actually on
// the line segment p1-p2
Point findIntersection(Point p1,Point p2,
Point p3,Point p4) {
float xD1,yD1,xD2,yD2,xD3,yD3;
float dot,deg,len1,len2;
float segmentLen1,segmentLen2;
float ua,ub,div;
// calculate differences
xD1=p2.x-p1.x;
xD2=p4.x-p3.x;
yD1=p2.y-p1.y;
yD2=p4.y-p3.y;
xD3=p1.x-p3.x;
yD3=p1.y-p3.y;
// calculate the lengths of the two lines
len1=sqrt(xD1*xD1+yD1*yD1);
len2=sqrt(xD2*xD2+yD2*yD2);
// calculate angle between the two lines.
dot=(xD1*xD2+yD1*yD2); // dot product
deg=dot/(len1*len2);
// if abs(angle)==1 then the lines are parallell,
// so no intersection is possible
if(abs(deg)==1) return null;
// find intersection Pt between two lines
Point pt=new Point(0,0);
div=yD2*xD1-xD2*yD1;
ua=(xD2*yD3-yD2*xD3)/div;
ub=(xD1*yD3-yD1*xD3)/div;
pt.x=p1.x+ua*xD1;
pt.y=p1.y+ua*yD1;
// calculate the combined length of the two segments
// between Pt-p1 and Pt-p2
xD1=pt.x-p1.x;
xD2=pt.x-p2.x;
yD1=pt.y-p1.y;
yD2=pt.y-p2.y;
segmentLen1=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);
// calculate the combined length of the two segments
// between Pt-p3 and Pt-p4
xD1=pt.x-p3.x;
xD2=pt.x-p4.x;
yD1=pt.y-p3.y;
yD2=pt.y-p4.y;
segmentLen2=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);
// if the lengths of both sets of segments are the same as
// the lenghts of the two lines the point is actually
// on the line segment.
// if the point isn’t on the line, return null
if(abs(len1-segmentLen1)>0.01 || abs(len2-segmentLen2)>0.01)
return null;
// return the valid intersection
return pt;
}
class Point{
float x,y;
Point(float x, float y){
this.x = x;
this.y = y;
}
void set(float x, float y){
this.x = x;
this.y = y;
}
}