a1,b1,c1是以fnode排序生成的數組，a1對應fnode,b1對應tnode,c1對應length,同樣a2,b2,c2,是以tnode 生成的數組。Indexa1是對應某壹起點與其相連的終點的個數，indexb1時對應某壹終點與其相連的起點的個數，即其拓撲關系。
Public Function shortpath(startno As Integer, endno As Integer) As Single
以開始點，結束點為參數。
Dim result() As Single
Dim result1 As Integer
定義結果點
Dim s1 As Single
Dim min As Single
Dim ii, i, j, aa As Integer
Dim yc() As Boolean
Dim ycd() As Boolean
Dim rs1() As Single
Dim no() As Integer
Dim nopoint As Integer
ReDim yc(1 To maxno) As Boolean
ReDim ycd(1 To maxno) As Boolean
ReDim rs1(1 To maxno) As Single
ReDim result(1 To 2, 1 To maxno) As Single
定義結果，其中result(1,maxno)為結果點，result(2,maxno)為結果長度。
For i = 1 To maxno// maxno為網中最大的節點數。
yc(i) = False //標記已經查過的點。
ycd(i) = False //標記已經作結果點用過的點
rs1(i) = 1E+38 //假設從起點到任壹點的距離都為無窮大
Next i
ll = startno //設置開始點。
yc(ll) = True //標記開始點為真。即已經作結果點用過。
j = 0
For aa = 1 To maxno
先從與開始點相連的終點尋找
For i = 1 To indexa1(2, ll) //以與ll點相連的起點的個數循環
result1 = b1(indexa1(1, ll) - i + 1)找出與LL點相連的終點的點號
s1 = c1(indexa1(1, ll) - i + 1) + result(2, ll)找出長度並求和
If yc(result1) = True Then GoTo 200如果以被經查過進行下壹個
If ycd(result1) = True Then//如果已經作為結果點判斷哪壹個長
If rs1(result1) >= s1 Then//如果這壹點到起點的長度比現在的路線長，替代
rs1(result1) = s1
result(1, result1) = ll//設置到這點的最短路徑的前壹點為LL點（精華部分）
result(2, result1) = s1設置到這點的最短路徑長度
GoTo 200
Else
GoTo 200
End If
End If
如果上面的條件都不符合則進行下面的語句
ycd(result1) = True
rs1(result1) = s1
result(1, result1) = ll
result(2, result1) = s1
每找到壹個點加壹，為了下面的判斷
j = j + 1
ReDim Preserve no(1 To j) As Integer
從新 定義數組並使其值為當前的點號
no(j) = result1
200 Next I
再從與開始點相連的終點尋找，與上面壹樣不再標註
For i = 1 To indexb2(2, ll)
result1 = a2(indexb2(1, ll) - i + 1)
s1 = c2(indexb2(1, ll) - i + 1) + result(2, ll)
If yc(result1) = True Then GoTo 300
If ycd(result1) = True Then
If rs1(result1) >= s1 Then
rs1(result1) = s1
result(1, result1) = ll
result(2, result1) = s1
GoTo 300
Else
GoTo 300
End If
End If
ycd(result1) = True
rs1(result1) = s1
result(1, result1) = ll
result(2, result1) = s1
j = j + 1
ReDim Preserve no(1 To j) As Integer
no(j) = result1
300 Next I
設置最小為無窮大，最短路徑點為空
min = 1E+38
minpoint = Null
（優化部分）
找出已經查過點中長度最短的點
For i = aa To j
If min > rs1(no(i)) Then
ii = i
min = rs1(no(i))
minpoint = no(i)
End If
Next I
如果沒有結果，即起點與終點沒有通路退出程序
If min = 1E+38 Then Exit Function
（重點優化）將兩點互換，減少循環。
no(ii) = no(aa)
no(aa) = minpoint
標記已經作為結果點判斷過
yc(minpoint) = True
ll = minpoint
判斷結果點是否等於終點，如果等於則已經找到最短路徑
If minpoint = endno Then Exit For
Next aa
返回最短路徑長度
Stpath = result(2, endno)
End Function