问题
In Dijkstra's algorithm what should I do if there are two or more nodes with minimal weights at a point in the algorithm?
In wikipedia: http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm at step no. 6, it says
"Set the unvisited node marked with the smallest tentative distance as the next 'current node' and go back to step 3."
What if there are two or more nodes with "smallest tentative distance".
Can anyone help me with the algorithm?
回答1:
Short Answer
Just pick either. Unless you have another heuristic to work with, you can't tell which would be better to pick.
Bit More Explaination
Think about sorting some elements into an array:
9 6 3 3 8
sorted with lowest first is
3 3 6 8 9
If you were to query that array to determine the lowest, the answer is 3. Which 3 doesn't matter.
Similarly if you were to have some more information. Say for example that those ints were actually floats and were sorted by their integer parts. You might end up with the array:
3.2 3.1 6.0 8.5 9.2
Here you have another heuristic to work with, you can check the decimal part too, and you can determine that 3.1 is the lowest.
来源：https://stackoverflow.com/questions/9264799/dijkstras-algorithm-what-to-do-if-there-are-two-or-more-nodes-with-minimal-wei