3d立方体网格的顶点c ++

时间:2013-07-11 10:39:54

标签: c++ 3d grid cube

我想在c ++中创建一个3d立方体网格,但我有一个问题:一旦指定起点(xs,ys,zs),终点(xe,ye,ze)和三个元素的数量axis..i想要识别每个立方体的八个角.. 设置原点在(0,0,0)和i,j,k中沿三轴移动的索引我知道:

1°角位于(i,j,k) 2°角位于(i + 1,j,k) 3°角位于(i,j + 1,k)

依旧......

我不知道如何获取这个三倍的价值,并在所有元素的三个“for循环”中识别出一个单点...... 请帮帮我:)。

1 个答案:

答案 0 :(得分:2)

首先定义一个简单的struct point3D:

typedef struct {

  float x;
  float y;
  float z;
} point3D;

我写这个来生成网格:

//Compute edge sizes 
float x_size = xe - xs;
float y_size = ye - ys;
float z_size = ze - zs;

//Compute steps
float x_step = x_size/n_step;
float y_step = y_size/n_step;
float z_step = z_size/n_step;

//Points per edge (including ending point)
int n = n_step + 1;

//Alloc grid (you can use malloc if you prefer)
point3D grid[n*n*n];

for (int i = 0; i < n; i++) { //move on x axis

  for (int j = 0; j < n; j++) { //move on y axis

    for (int k = 0; k < n; k++) { //move on z axis

      point3D p;
      p.x = xs + x_step * i;
      p.y = ys + y_step * j;
      p.z = zs + z_step * k;
      grid[i+n*j+n*n*k] = point3D;
    } 
  }
}

使用8角点:

point3D corner = grid[n_step*x + n*n_step*y + n*n*n_step*z];

使用:

   (x, y, z)
1: (0, 0, 0)
2: (0, 0, 1)
3: (0, 1, 0)
4: (0, 1, 1)
5: (1, 0, 0)
6: (1, 0, 1)
7: (1, 1, 0)
8: (1, 1, 1)