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
这是确定一个点是在三角形的内、外还是在三角形的臂上的最简单的概念。
用行列式确定三角形内的点:
最简单的工作代码:
#-*- coding: utf-8 -*-
import numpy as np
tri_points = [(1,1),(2,3),(3,1)]
def pisinTri(point,tri_points):
Dx , Dy = point
A,B,C = tri_points
Ax, Ay = A
Bx, By = B
Cx, Cy = C
M1 = np.array([ [Dx - Bx, Dy - By, 0],
[Ax - Bx, Ay - By, 0],
[1 , 1 , 1]
])
M2 = np.array([ [Dx - Ax, Dy - Ay, 0],
[Cx - Ax, Cy - Ay, 0],
[1 , 1 , 1]
])
M3 = np.array([ [Dx - Cx, Dy - Cy, 0],
[Bx - Cx, By - Cy, 0],
[1 , 1 , 1]
])
M1 = np.linalg.det(M1)
M2 = np.linalg.det(M2)
M3 = np.linalg.det(M3)
print(M1,M2,M3)
if(M1 == 0 or M2 == 0 or M3 ==0):
print("Point: ",point," lies on the arms of Triangle")
elif((M1 > 0 and M2 > 0 and M3 > 0)or(M1 < 0 and M2 < 0 and M3 < 0)):
#if products is non 0 check if all of their sign is same
print("Point: ",point," lies inside the Triangle")
else:
print("Point: ",point," lies outside the Triangle")
print("Vertices of Triangle: ",tri_points)
points = [(0,0),(1,1),(2,3),(3,1),(2,2),(4,4),(1,0),(0,4)]
for c in points:
pisinTri(c,tri_points)
其他回答
一般来说，最简单(也是最优)的算法是检查由边创建的半平面的哪一边是点。
以下是关于GameDev的一些高质量信息，包括性能问题。
这里有一些代码让你开始:
float sign (fPoint p1, fPoint p2, fPoint p3)
{
return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}
bool PointInTriangle (fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
float d1, d2, d3;
bool has_neg, has_pos;
d1 = sign(pt, v1, v2);
d2 = sign(pt, v2, v3);
d3 = sign(pt, v3, v1);
has_neg = (d1 < 0) || (d2 < 0) || (d3 < 0);
has_pos = (d1 > 0) || (d2 > 0) || (d3 > 0);
return !(has_neg && has_pos);
}
由andreasdr和Perro Azul发布的重心方法的c#版本。我添加了一个检查，当s和t有相反的符号(而且都不为零)时，放弃面积计算，因为潜在地避免三分之一的乘法成本似乎是合理的。
public static bool PointInTriangle(Point p, Point p0, Point p1, Point p2)
{
var s = (p0.X - p2.X) * (p.Y - p2.Y) - (p0.Y - p2.Y) * (p.X - p2.X);
var t = (p1.X - p0.X) * (p.Y - p0.Y) - (p1.Y - p0.Y) * (p.X - p0.X);
if ((s < 0) != (t < 0) && s != 0 && t != 0)
return false;
var d = (p2.X - p1.X) * (p.Y - p1.Y) - (p2.Y - p1.Y) * (p.X - p1.X);
return d == 0 || (d < 0) == (s + t <= 0);
}
2021年更新:这个版本正确处理任意一个缠绕方向(顺时针和逆时针)指定的三角形。请注意，对于恰好位于三角形边缘上的点，本页上的一些其他答案会给出不一致的结果，这取决于三角形三个点的排列顺序。这些点被认为是“在”三角形中，这段代码正确地返回true，而不管缠绕方向如何。
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
bool point2Dtriangle(double e,double f, double a,double b,double c, double g,double h,double i, double v, double w){
/* inputs: e=point.x, f=point.y
a=triangle.Ax, b=triangle.Bx, c=triangle.Cx
g=triangle.Ay, h=triangle.By, i=triangle.Cy */
v = 1 - (f * (b - c) + h * (c - e) + i * (e - b)) / (g * (b - c) + h * (c - a) + i * (a - b));
w = (f * (a - b) + g * (b - e) + h * (e - a)) / (g * (b - c) + h * (c - a) + i * (a - b));
if (*v > -0.0 && *v < 1.0000001 && *w > -0.0 && *w < *v) return true;//is inside
else return false;//is outside
return 0;
}
从质心转换而来的几乎完美的笛卡尔坐标 在*v (x)和*w (y)双精度内导出。 在每种情况下，两个导出双精度对象前面都应该有一个*字符，可能是*v和*w 代码也可以用于四边形的另一个三角形。 特此签名只写三角形abc从顺时针abcd的四边形。
A---B
|..\\.o|
|....\\.|
D---C
o点在ABC三角形内 对于带有第二个三角形的测试，将此函数称为CDA方向，*v=1-*v后的结果应正确;* w = 1 - * w;为了四合院
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
