Question

我正在尝试创建UV @运行时，我正在使用BOX类型的UV（类似于3ds max中的BOX UVW）并基于面部方向的计算。

我知道将它创建为运行时不是一个好选择，但我别无选择:(它在计算后保存，所以我做了一次。

但我需要40秒才能获得30000个顶点......太长了

我的代码中是否有任何可以优化的优化？

如果你有＆lt; 5000 vertices Mesh：

，我的代码可以随意使用

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。

vertices属性的文档部分说明：

返回顶点位置的副本或指定新的顶点位置数组。

注意“返回副本” - 您在循环内访问此属性22次，因此这将创建该数组的大约22n / 3个副本。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（就像您已经使用mesh.uvs一样），您应该会看到显着的性能提升。

您还可以检查mesh.triangles是否为复制操作。我预计它可能会，但文档没有指定。

Answer 2

这是我的代码优化，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;
}

UV解包运行时优化

2 个答案: