Hi everyone ...i would like your ideas on how to go about this list problem.
attached is a sample of the list am trying to work with:
I want to achieve a list of possible route like this
possible route 1 - 272<309<308>312>48
route 2 - 316<315>319>99
route 3 - 48<1420
< it begins from the left of the node
> it end at the right side of the node
any help ?
Path node , route path, list algorithm
Okay Guys thanks very much i noticed i had over 100 views yet no one came to my rescue.
Well i was able to figure it out. in case anyone chanced upon this search and finds a simlar situation this is how
i went about it. Using Graphs Representation and BFS & DFS for path or route finding
for now that is the approach i have used and seems to work well
Well i was able to figure it out. in case anyone chanced upon this search and finds a simlar situation this is how
i went about it. Using Graphs Representation and BFS & DFS for path or route finding
- i represented each conn.# or connector.# as a vertices or a node
- created an undirected edge as a link or a connection to another conn.# object
- created a graph out of the table
- Used BFS,DFS to find the paths or route.
for now that is the approach i have used and seems to work well
Can the component graph have a loop? That would be something you need to take into account.
FYI: This happens to be the first time I've come across this thread.
That usually means that you haven't explained your question very well, or for some other reason it is going to require too much of our time to understand and help with.
The easier you make it for others to help you, the more help you'll get, and faster. That's just how it is.
Okay Guys thanks very much i noticed i had over 100 views yet no one came to my rescue.
That usually means that you haven't explained your question very well, or for some other reason it is going to require too much of our time to understand and help with.
The easier you make it for others to help you, the more help you'll get, and faster. That's just how it is.
This topic is closed to new replies.
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