本文整理汇总了C#中Vertices.Clear方法的典型用法代码示例。如果您正苦于以下问题：C# Vertices.Clear方法的具体用法？C# Vertices.Clear怎么用？C# Vertices.Clear使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Vertices的用法示例。
在下文中一共展示了Vertices.Clear方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: MergeParallelEdges
//From Eric Jordan's convex decomposition library
/// <summary>
/// Merges all parallel edges in the list of vertices
/// </summary>
/// <param name="vertices">The vertices.</param>
/// <param name="tolerance">The tolerance.</param>
public static void MergeParallelEdges(Vertices vertices, float tolerance)
{
if (vertices.Count <= 3)
return; //Can't do anything useful here to a triangle
bool[] mergeMe = new bool[vertices.Count];
int newNVertices = vertices.Count;
//Gather points to process
for (int i = 0; i < vertices.Count; ++i)
{
int lower = (i == 0) ? (vertices.Count - 1) : (i - 1);
int middle = i;
int upper = (i == vertices.Count - 1) ? (0) : (i + 1);
float dx0 = vertices[middle].X - vertices[lower].X;
float dy0 = vertices[middle].Y - vertices[lower].Y;
float dx1 = vertices[upper].Y - vertices[middle].X;
float dy1 = vertices[upper].Y - vertices[middle].Y;
float norm0 = (float) Math.Sqrt(dx0*dx0 + dy0*dy0);
float norm1 = (float) Math.Sqrt(dx1*dx1 + dy1*dy1);
if (!(norm0 > 0.0f && norm1 > 0.0f) && newNVertices > 3)
{
//Merge identical points
mergeMe[i] = true;
--newNVertices;
}
dx0 /= norm0;
dy0 /= norm0;
dx1 /= norm1;
dy1 /= norm1;
float cross = dx0*dy1 - dx1*dy0;
float dot = dx0*dx1 + dy0*dy1;
if (Math.Abs(cross) < tolerance && dot > 0 && newNVertices > 3)
{
mergeMe[i] = true;
--newNVertices;
}
else
mergeMe[i] = false;
}
if (newNVertices == vertices.Count || newNVertices == 0)
return;
int currIndex = 0;
//Copy the vertices to a new list and clear the old
Vertices oldVertices = new Vertices(vertices);
vertices.Clear();
for (int i = 0; i < oldVertices.Count; ++i)
{
if (mergeMe[i] || newNVertices == 0 || currIndex == newNVertices)
continue;
Debug.Assert(currIndex < newNVertices);
vertices.Add(oldVertices[i]);
++currIndex;
}
}