有没有一种简单的方法来确定一个点是否在三角形内?是2D的,不是3D的。
当前回答
我在最后一次尝试谷歌和找到这个页面之前写了这段代码,所以我想我应该分享它。它基本上是Kisielewicz答案的优化版本。我也研究了重心法,但从维基百科的文章来看,我很难看出它是如何更有效的(我猜有一些更深层次的等价性)。不管怎样,这个算法的优点是不用除法;一个潜在的问题是边缘检测的行为取决于方向。
bool intpoint_inside_trigon(intPoint s, intPoint a, intPoint b, intPoint c)
{
int as_x = s.x - a.x;
int as_y = s.y - a.y;
bool s_ab = (b.x - a.x) * as_y - (b.y - a.y) * as_x > 0;
if ((c.x - a.x) * as_y - (c.y - a.y) * as_x > 0 == s_ab)
return false;
if ((c.x - b.x) * (s.y - b.y) - (c.y - b.y)*(s.x - b.x) > 0 != s_ab)
return false;
return true;
}
换句话说,思想是这样的:点s是在直线AB和直线AC的左边还是右边?如果是真的,它就不可能在里面。如果为假,则至少在“锥”内满足条件。现在,因为我们知道三角形(三角形)内的一个点必须与BC(以及CA)在AB的同一侧,我们检查它们是否不同。如果有,s就不可能在里面,否则s一定在里面。
计算中的一些关键字是线半平面和行列式(2x2叉乘)。也许一个更有教学意义的方法是将它看作是一个在AB、BC和CA的同一侧(左或右)的点。然而,上面的方法似乎更适合进行一些优化。
其他回答
我在JavaScript中改编的高性能代码(文章如下):
function pointInTriangle (p, p0, p1, p2) {
return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0;
}
pointInTriangle(p, p0, p1, p2) -用于逆时针方向的三角形 pointInTriangle(p, p0, p1, p2) -用于顺时针三角形
在jsFiddle(包括性能测试)中,在一个单独的函数中也有缠绕检查。或按下面的“运行代码片段”
var ctx = $("canvas")[0].getContext("2d"); var W = 500; var H = 500; var point = { x: W / 2, y: H / 2 }; var triangle = randomTriangle(); $("canvas").click(function(evt) { point.x = evt.pageX - $(this).offset().left; point.y = evt.pageY - $(this).offset().top; test(); }); $("canvas").dblclick(function(evt) { triangle = randomTriangle(); test(); }); document.querySelector('#performance').addEventListener('click', _testPerformance); test(); function test() { var result = checkClockwise(triangle.a, triangle.b, triangle.c) ? pointInTriangle(point, triangle.a, triangle.c, triangle.b) : pointInTriangle(point, triangle.a, triangle.b, triangle.c); var info = "point = (" + point.x + "," + point.y + ")\n"; info += "triangle.a = (" + triangle.a.x + "," + triangle.a.y + ")\n"; info += "triangle.b = (" + triangle.b.x + "," + triangle.b.y + ")\n"; info += "triangle.c = (" + triangle.c.x + "," + triangle.c.y + ")\n"; info += "result = " + (result ? "true" : "false"); $("#result").text(info); render(); } function _testPerformance () { var px = [], py = [], p0x = [], p0y = [], p1x = [], p1y = [], p2x = [], p2y = [], p = [], p0 = [], p1 = [], p2 = []; for(var i = 0; i < 1000000; i++) { p[i] = {x: Math.random() * 100, y: Math.random() * 100}; p0[i] = {x: Math.random() * 100, y: Math.random() * 100}; p1[i] = {x: Math.random() * 100, y: Math.random() * 100}; p2[i] = {x: Math.random() * 100, y: Math.random() * 100}; } console.time('optimal: pointInTriangle'); for(var i = 0; i < 1000000; i++) { pointInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('optimal: pointInTriangle'); console.time('original: ptInTriangle'); for(var i = 0; i < 1000000; i++) { ptInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('original: ptInTriangle'); } function pointInTriangle (p, p0, p1, p2) { return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0; } function ptInTriangle(p, p0, p1, p2) { var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y); var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y); if (s <= 0 || t <= 0) return false; var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return (s + t) < A; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function checkClockwise(p0, p1, p2) { var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return A > 0; } function drawTriangle(p0, p1, p2) { ctx.fillStyle = "#999"; ctx.beginPath(); ctx.moveTo(p0.x, p0.y); ctx.lineTo(p1.x, p1.y); ctx.lineTo(p2.x, p2.y); ctx.closePath(); ctx.fill(); ctx.fillStyle = "#000"; ctx.font = "12px monospace"; ctx.fillText("1", p0.x, p0.y); ctx.fillText("2", p1.x, p1.y); ctx.fillText("3", p2.x, p2.y); } function drawPoint(p) { ctx.fillStyle = "#F00"; ctx.beginPath(); ctx.arc(p.x, p.y, 5, 0, 2 * Math.PI); ctx.fill(); } function rand(min, max) { return Math.floor(Math.random() * (max - min + 1)) + min; } function randomTriangle() { return { a: { x: rand(0, W), y: rand(0, H) }, b: { x: rand(0, W), y: rand(0, H) }, c: { x: rand(0, W), y: rand(0, H) } }; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <button id="performance">Run performance test (open console)</button> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>
受此启发: http://www.phatcode.net/articles.php?id=459
If you know the co-ordinates of the three vertices and the co-ordinates of the specific point, then you can get the area of the complete triangle. Afterwards, calculate the area of the three triangle segments (one point being the point given and the other two being any two vertices of the triangle). Thus, you will get the area of the three triangle segments. If the sum of these areas are equal to the total area (that you got previously), then, the point should be inside the triangle. Otherwise, the point is not inside the triangle. This should work. If there are any issues, let me know. Thank you.
我同意Andreas Brinck的观点,重心坐标对于这项任务来说非常方便。注意,不需要每次都求解一个方程组:只需计算解析解。使用Andreas的符号,解是:
s = 1/(2*Area)*(p0y*p2x - p0x*p2y + (p2y - p0y)*px + (p0x - p2x)*py);
t = 1/(2*Area)*(p0x*p1y - p0y*p1x + (p0y - p1y)*px + (p1x - p0x)*py);
其中Area是三角形的(带符号的)面积:
Area = 0.5 *(-p1y*p2x + p0y*(-p1x + p2x) + p0x*(p1y - p2y) + p1x*p2y);
只计算st和1-s-t。点p在三角形内当且仅当它们都是正的。
编辑:请注意,上面的区域表达式假设三角形节点编号是逆时针方向的。如果编号是顺时针的,这个表达式将返回一个负的面积(但大小正确)。然而,测试本身(s>0 && t>0 && 1-s-t>0)并不依赖于编号的方向,因为如果三角形节点的方向改变,上面乘以1/(2*Area)的表达式也会改变符号。
编辑2:为了获得更好的计算效率,请参阅下面的coproc注释(其中指出,如果三角形节点的方向(顺时针或逆时针)事先已知,则可以避免在s和t的表达式中除以2*Area)。在Andreas Brinck的回答下面的评论中也可以看到Perro Azul的jsfiddle-code。
重心法Java版:
class Triangle {
Triangle(double x1, double y1, double x2, double y2, double x3,
double y3) {
this.x3 = x3;
this.y3 = y3;
y23 = y2 - y3;
x32 = x3 - x2;
y31 = y3 - y1;
x13 = x1 - x3;
det = y23 * x13 - x32 * y31;
minD = Math.min(det, 0);
maxD = Math.max(det, 0);
}
boolean contains(double x, double y) {
double dx = x - x3;
double dy = y - y3;
double a = y23 * dx + x32 * dy;
if (a < minD || a > maxD)
return false;
double b = y31 * dx + x13 * dy;
if (b < minD || b > maxD)
return false;
double c = det - a - b;
if (c < minD || c > maxD)
return false;
return true;
}
private final double x3, y3;
private final double y23, x32, y31, x13;
private final double det, minD, maxD;
}
上面的代码可以准确地处理整数,假设没有溢出。它也适用于顺时针和逆时针三角形。它不适用于共线三角形(但您可以通过测试det==0来检查)。
如果你要用同一个三角形测试不同的点,以重心为中心的版本是最快的。
重心版本在3个三角形点上是不对称的,所以它可能不如Kornel Kisielewicz的边缘半平面版本一致,因为浮点舍入误差。
图片来源:我根据维基百科关于重心坐标的文章制作了上面的代码。
最简单的方法,适用于所有类型的三角形,就是确定P点A点B点C点的角。如果任何一个角大于180.0度,那么它在外面,如果是180.0度,那么它在圆周上,如果acos欺骗了你,小于180.0度,那么它在里面。看一看理解http://math-physics-psychology.blogspot.hu/2015/01/earlish-determination-that-point-is.html