翻转两个三角形的边

这取决于你如何实现你的结构。例如,如果您有这样一个结构:

struct Vertex
{
    double position[3];
};

struct Triangle
{
    unsigned int vertices[3];
};

Vertex nodes[N_NODES];
Triangle triangles[N_TRIANGLES];

其中, vertices 是的索引 nodes ,则给定两个三角形:

Triangle t1, t2;

如果你有

t1.vertices[i1] = t2.vertices[j1];
t1.vertices[i2] = t2.vertices[j2];

这意味着 t1.vertices[i1] --- t1.vertices[i2] 是与共享的边缘 t2.vertices[j1] --- t1.vertices[j2] ,并假设 i3 和 j3 是的第三个真理 t1 和 t2 如该图像所示:

然后,只需更改一些索引就可以进行翻转。最直接的方法是:

unsigned int t1_v[3], t2_v[3];

t1_v[0] = t1.vertices[i1];  // shared
t1_v[1] = t1.vertices[i2];  // shared
t1_v[2] = t1.vertices[i3];  // not shared

t2_v[0] = t2.vertices[j1];  // shared (unnecessary)
t2_v[1] = t2.vertices[j2];  // shared (unnecessary)
t2_v[2] = t2.vertices[j3];  // not shared

t1.vertices[0] = t1_v[0];  // previously shared
t1.vertices[1] = t1_v[2];  // previously not shared
t1.vertices[2] = t2_v[2];  // previously not shared

t2.vertices[0] = t2_v[0];  // previously shared
t2.vertices[1] = t2_v[2];  // previously not shared
t2.vertices[2] = t1_v[2];  // previously not shared

请注意,每个新三角形由2个非共享顶点和一个共享顶点组成。

如果你的顶点在一个数组中,我假设你通过一些固定的行长度识别不同的行,并根据一些模式制作三角形。我建议只是改变这种模式,如何将三个顶点指定在一起。你必须注意没有配对的三角形,如果有