我試圖確定一個(gè)點(diǎn)是否在多邊形內(nèi)。多邊形是由Point對(duì)象數(shù)組定義的。我可以很容易地弄清楚該點(diǎn)是否在多邊形的邊界框內(nèi)，但是我不確定如何判斷它是否在實(shí)際的多邊形內(nèi)。如果可能的話，我只想使用C＃和WinForms。我寧愿不調(diào)用OpenGL或執(zhí)行某些簡(jiǎn)單任務(wù)。這是我到目前為止的代碼：private void CalculateOuterBounds(){ //m_aptVertices is a Point[] which holds the vertices of the polygon. // and X/Y min/max are just ints Xmin = Xmax = m_aptVertices[0].X; Ymin = Ymax = m_aptVertices[0].Y; foreach(Point pt in m_aptVertices) { if(Xmin > pt.X) Xmin = pt.X; if(Xmax < pt.X) Xmax = pt.X; if(Ymin > pt.Y) Ymin = pt.Y; if(Ymax < pt.Y) Ymax = pt.Y; }}public bool Contains(Point pt){ bool bContains = true; //obviously wrong at the moment :) if(pt.X < Xmin || pt.X > Xmax || pt.Y < Ymin || pt.Y > Ymax) bContains = false; else { //figure out if the point is in the polygon } return bContains;}
3 回答
RISEBY
TA貢獻(xiàn)1856條經(jīng)驗(yàn) 獲得超5個(gè)贊
我在這里檢查了代碼，都遇到了問(wèn)題。
最好的方法是：
/// <summary>
/// Determines if the given point is inside the polygon
/// </summary>
/// <param name="polygon">the vertices of polygon</param>
/// <param name="testPoint">the given point</param>
/// <returns>true if the point is inside the polygon; otherwise, false</returns>
public static bool IsPointInPolygon4(PointF[] polygon, PointF testPoint)
{
bool result = false;
int j = polygon.Count() - 1;
for (int i = 0; i < polygon.Count(); i++)
{
if (polygon[i].Y < testPoint.Y && polygon[j].Y >= testPoint.Y || polygon[j].Y < testPoint.Y && polygon[i].Y >= testPoint.Y)
{
if (polygon[i].X + (testPoint.Y - polygon[i].Y) / (polygon[j].Y - polygon[i].Y) * (polygon[j].X - polygon[i].X) < testPoint.X)
{
result = !result;
}
}
j = i;
}
return result;
}
蕪湖不蕪
TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超7個(gè)贊
接受的答案對(duì)我的項(xiàng)目不起作用。我最終使用了在這里找到的代碼。
public static bool IsInPolygon(Point[] poly, Point p)
{
Point p1, p2;
bool inside = false;
if (poly.Length < 3)
{
return inside;
}
var oldPoint = new Point(
poly[poly.Length - 1].X, poly[poly.Length - 1].Y);
for (int i = 0; i < poly.Length; i++)
{
var newPoint = new Point(poly[i].X, poly[i].Y);
if (newPoint.X > oldPoint.X)
{
p1 = oldPoint;
p2 = newPoint;
}
else
{
p1 = newPoint;
p2 = oldPoint;
}
if ((newPoint.X < p.X) == (p.X <= oldPoint.X)
&& (p.Y - (long) p1.Y)*(p2.X - p1.X)
< (p2.Y - (long) p1.Y)*(p.X - p1.X))
{
inside = !inside;
}
oldPoint = newPoint;
}
return inside;
}
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