我创建了一个名为 isInPoly 的方法,它接收 2 条线的 x 和 y 坐标(所以 8 个坐标)并确定 2 条线相交的位置。我知道我应该调用类似 intersectLocation 之类的方法,但创建它的原因是查看一个点是否在多边形中。如果您知道一个不在多边形中的点,并在该点和您要测试的点之间划一条线以查看它是否在多边形内,然后计算有多少交叉点。如果交叉点的数量是偶数,则该点不在多边形中,如果数量为奇数,则该点在多边形中。无论如何,我没有得到这种方法的正确输出。如果测试点与已知点的斜率为 -1,我的程序仅向我显示该点位于多边形内部。
public static boolean isInPoly(float l1x1, float l1y1, float l1x2, float l1y2, float l2x1, float l2y1, float l2x2, float l2y2) {
// TODO Auto-generated method stub
// l1x1 = the first lines first x coordinate
// l1y1 = the first lines first x coordinate
//.....
//l1m = the first lines slope represented as "m" in the equation y=mx+b
//l1b = the first lines y intercept represented as "b" in the equation y=mx+b
// x = the x coordinate of the intersection on the 2 lines
// y = the x coordinate of the intersection on the 2 lines
float l1m,l2m,l1b,l2b,x,y;
//y=mx+b
//x=(y2-y1)/(x2-x1)
//b=y/(mx)
//slopes of each line
l1m = (l1y2-l1y1)/(l1x2-l1x1);
l2m = (l2y2-l2y1)/(l2x2-l2x1);
//y-intercepts of each line
l1b = l1y2/(l1m*l1x2);
l2b = l2y2/(l2m*l2x2);
//m1x+b1=m2x+b2
//m1x=m2x+b2-b1
//x=(m2/m1)x+((b2-b1)/m1)
//x-(m2/m1)x=((b2-b1)/m1)
//(1-(m2/m1))x=((b2-b1)/m1)
//x=((b2-b1)/m1)/(1-(m2/m1))
//finding the x coordinate of the intersection
x=((l2b-l1b)/l1m)/(1-(l2m/l1m));
//y=mx+b
//finding the x coordinate of the intersection
y=(l1m*x)+l1b;
if(y>=l1y1 && y<=l1y2 && x>=l1x1 && x<=l1x2){
return true;
}
else{
return false;
}
}