运行时优化的UV展开

Question

我试图在运行时创建UVs，我使用BOX类型的UV（类似于3ds max中的BOX UVW），并基于我的面部方向计算。

我知道这不是一个很好的选择，以创建一个运行时间，但我别无选择:(它的计算后保存的，所以我做了一次。

但我需要40秒，一个30000点的顶点......太长

有没有在我的代码任何优化，可以进行

这里是我的代码随意使用，如果你有< 5000顶点网：？

public static void CreateUV(ref Mesh mesh) 
{ 

    int i = 0; 
    Vector3 p = Vector3.up; 
    Vector3 u = Vector3.Cross(p, Vector3.forward); 
    if (Vector3.Dot(u, u) < 0.001f) 
    { 
     u = Vector3.right; 
    } 
    else 
    { 
     u = Vector3.Normalize(u); 
    } 

    Vector3 v = Vector3.Normalize(Vector3.Cross(p, u)); 

    Vector2[] uvs = new Vector2[mesh.vertices.Length]; 

    for (i = 0; i < mesh.triangles.Length; i += 3) 
    { 

     Vector3 a = mesh.vertices[mesh.triangles[i]]; 
     Vector3 b = mesh.vertices[mesh.triangles[i + 1]]; 
     Vector3 c = mesh.vertices[mesh.triangles[i + 2]]; 
     Vector3 side1 = b - a; 
     Vector3 side2 = c - a; 
     Vector3 N = Vector3.Cross(side1, side2); 

     N = new Vector3(Mathf.Abs(N.normalized.x), Mathf.Abs(N.normalized.y), Mathf.Abs(N.normalized.z)); 



     if (N.x > N.y && N.x > N.z) 
     { 
      uvs[mesh.triangles[i]] = new Vector2(mesh.vertices[mesh.triangles[i]].z, mesh.vertices[mesh.triangles[i]].y); 
      uvs[mesh.triangles[i + 1]] = new Vector2(mesh.vertices[mesh.triangles[i + 1]].z, mesh.vertices[mesh.triangles[i + 1]].y); 
      uvs[mesh.triangles[i + 2]] = new Vector2(mesh.vertices[mesh.triangles[i + 2]].z, mesh.vertices[mesh.triangles[i + 2]].y); 
     } 
     else if (N.y > N.x && N.y > N.z) 
     { 
      uvs[mesh.triangles[i]] = new Vector2(mesh.vertices[mesh.triangles[i]].x, mesh.vertices[mesh.triangles[i]].z); 
      uvs[mesh.triangles[i + 1]] = new Vector2(mesh.vertices[mesh.triangles[i + 1]].x, mesh.vertices[mesh.triangles[i + 1]].z); 
      uvs[mesh.triangles[i + 2]] = new Vector2(mesh.vertices[mesh.triangles[i + 2]].x, mesh.vertices[mesh.triangles[i + 2]].z); 
     } 
     else if (N.z > N.x && N.z > N.y) 
     { 
      uvs[mesh.triangles[i]] = new Vector2(mesh.vertices[mesh.triangles[i]].x, mesh.vertices[mesh.triangles[i]].y); 
      uvs[mesh.triangles[i + 1]] = new Vector2(mesh.vertices[mesh.triangles[i + 1]].x, mesh.vertices[mesh.triangles[i + 1]].y); 
      uvs[mesh.triangles[i + 2]] = new Vector2(mesh.vertices[mesh.triangles[i + 2]].x, mesh.vertices[mesh.triangles[i + 2]].y); 
     } 

    } 

    mesh.uv = uvs; 
    Debug.Log("Finish"); 
} 

Answer 1

我强烈建议缓存mesh.vertices的副本。

部分为vertices属性状态的文档：

返回顶点位置的副本或分配新的顶点位置阵列。

注“返回副本” - 你访问这个属性22倍的循环中，所以这将创建数组的大致22n/3副本。对于具有30,000个顶点的网格，这是在后台进行的超过200,000次不必要的复制操作。

如果您创建一个临时数组来保存顶点数据（就像您已经在使用mesh.uvs一样），您应该看到显着的性能提升。

您可能还会检查mesh.triangles是否是复制操作。我预计它可能是，但文件没有具体说明。

Answer 2

这里是我的代码优化感谢@rutter

public static Vector2[] CreateUV(ref Mesh mesh) 
{ 

    int i = 0; 
    Vector3 p = Vector3.up; 
    Vector3 u = Vector3.Cross(p, Vector3.forward); 
    if (Vector3.Dot(u, u) < 0.001f) 
    { 
     u = Vector3.right; 
    } 
    else 
    { 
     u = Vector3.Normalize(u); 
    } 

    Vector3 v = Vector3.Normalize(Vector3.Cross(p, u)); 
    Vector3[] vertexs = mesh.vertices; 
    int[] tris = mesh.triangles; 
    Vector2[] uvs = new Vector2[vertexs.Length]; 

    for (i = 0; i < tris.Length; i += 3) 
    { 

     Vector3 a = vertexs[tris[i]]; 
     Vector3 b = vertexs[tris[i + 1]]; 
     Vector3 c = vertexs[tris[i + 2]]; 
     Vector3 side1 = b - a; 
     Vector3 side2 = c - a; 
     Vector3 N = Vector3.Cross(side1, side2); 

     N = new Vector3(Mathf.Abs(N.normalized.x), Mathf.Abs(N.normalized.y), Mathf.Abs(N.normalized.z)); 



     if (N.x > N.y && N.x > N.z) 
     { 
      uvs[tris[i]] = new Vector2(vertexs[tris[i]].z, vertexs[tris[i]].y); 
      uvs[tris[i + 1]] = new Vector2(vertexs[tris[i + 1]].z, vertexs[tris[i + 1]].y); 
      uvs[tris[i + 2]] = new Vector2(vertexs[tris[i + 2]].z, vertexs[tris[i + 2]].y); 
     } 
     else if (N.y > N.x && N.y > N.z) 
     { 
      uvs[tris[i]] = new Vector2(vertexs[tris[i]].x, vertexs[tris[i]].z); 
      uvs[tris[i + 1]] = new Vector2(vertexs[tris[i + 1]].x, vertexs[tris[i + 1]].z); 
      uvs[tris[i + 2]] = new Vector2(vertexs[tris[i + 2]].x, vertexs[tris[i + 2]].z); 
     } 
     else if (N.z > N.x && N.z > N.y) 
     { 
      uvs[tris[i]] = new Vector2(vertexs[tris[i]].x, vertexs[tris[i]].y); 
      uvs[tris[i + 1]] = new Vector2(vertexs[tris[i + 1]].x, vertexs[tris[i + 1]].y); 
      uvs[tris[i + 2]] = new Vector2(vertexs[tris[i + 2]].x, vertexs[tris[i + 2]].y); 
     } 

    } 

    mesh.uv = uvs; 
    return uvs; 
}