[求助]判断点在闭合区域内部，代码有问题，高手帮忙

misters

'一个矩形的形状，用以下表示

dim str as string

str="30,-70|50,-70|50,-84|30,-84"

dim jl as long '输入一个值，在str矩形的内部画一个xy坐标都内移jl值的矩形，这个矩形正好套在上个乱形之内。

新得到的矩形的坐标值是"32,-72|48,-72|48,-82|32,-82"

我用的方法是将每个顶点可能发生的四种情况都调用ptinpoly函数判断如果返回0值表示在内部，但是四个值一个都不返回0值，下面的函数是不是有错误啊，高手能不能帮我检查一下，看看有什么问题，或是谁有VB或VBA的代码发表一下，急呀，检查不出问题呀。

它的四种情况是：

x-,y-

x+,y+

x-,y+

x+,y-

这四种情况中应该有一种是在内部的

'判断给定点 pt 是否在多边形 poly 内
'返回 0 在内部，-1 在外面
'返回 > 0 表示点在第几条有向线段上
Private Function PtInPoly(ByVal pt As Variant, ByVal poly As Variant) As Integer
    Dim i As Integer
    Dim status, lastStauts As Integer
    Dim cnt As Integer
    Dim POS, temp As Integer
   
    Dim var_poly As Variant
    Dim var_polyi As Variant
   
    var_poly = Split(poly(1), ",")
    cnt = 0
    lastStauts = IIf(var_poly(1) > pt(1), 1, IIf(var_poly(1) = pt(1), 0, -1))
   
    For i = 1 To UBound(poly)
        var_poly = Split(poly(i), ",")
        status = IIf(var_poly(1) > pt(1), 1, IIf(var_poly(1) < pt(1), -1, 0))
        temp = status - lastStauts
        lastStauts = status
        POS = SideOfLine(poly(i - 1), poly(i), pt)
'       点在有向线段上
        var_poly = Split(poly(i), ",")
        var_polyi = Split(poly(i - 1), ",")
        If (POS = 0 And (var_polyi(0) <= pt(0) And pt(0) <= var_poly(0) Or var_polyi(0) >= pt(0) And pt(0) >= var_poly(0)) And (var_polyi(1) <= pt(1) And pt(1) <= var_poly(1) Or var_polyi(1) >= pt(1) And pt(1) >= var_poly(1))) Then
            PtInPoly = i
            Exit Function
        End If
       
'        跨越
        If (temp > 0 And POS = 1 Or temp < 0 And POS = -1) Then
            cnt = cnt + temp
        End If
       
    Next
   
    PtInPoly = IIf(cnt = 0, -1, 0)
   
End Function

'判断点在线的哪侧
Private Function SideOfLine(ByVal p1 As Variant, ByVal p2 As Variant, ByVal pt As Variant) As Integer
    Dim RR, TOP, LL As Integer
    RR = -1
    TOP = 0
    LL = 1
   
    Dim c1 As Double
    Dim c2 As Double
    Dim var_p1 As Variant
    Dim var_p2 As Variant
   
    var_p1 = Split(p1, ",")
    var_p2 = Split(p2, ",")
   
    c1 = (var_p2(0) - pt(0)) * (pt(1) - var_p1(1))
    c2 = (var_p2(1) - pt(1)) * (pt(0) - var_p1(0))
    SideOfLine = IIf(c1 > c2, LL, IIf(c1 < c2, RR, TOP))
End Function