问题
I'm looking for an algorithm(C/C++/Java - doesn't matter) which will resolve a problem which consists of finding the shortest path between 2 nodes (A and B) of a graph. The catch is that the path must visit certain other given nodes(cities). A city can be visited more than once. Example of path(A-H-D-C-E-F-G-F-B) (where A is source, B is destination, F and G are cities which must be visited).
I see this as a variation of the Traveling Salesman Problem but I couldn't find or write a working algorithm based on my searches.
I was trying to find a solution starting from these topics but without any luck: https://stackoverflow.com/questions/24856875/tsp-branch-and-bound-implementation-in-c and Variation of TSP which visits multiple cities
回答1:
An easy reduction of the problem to TSP would be:
- For each
(u,v)that "must be visited", find the distanced(u,v)between them. This can be done efficiently using Floyd-Warshall Algorithm to find all-to-all shortest path. - Create a new graph G' consisting only of those nodes, with all edges existing, with the distances as calculated on (1).
- Run standard TSP algorithm to solve the problem on the reduced graph.
回答2:
I think that in addition to amit's answer, you'll want to increase the cost of the edges that have A or B as endpoints by a sufficient amount (the total cost of the graph + 1 would probably be sufficient) to ensure that you don't end up with a path that goes through A or B (instead of ending at A and B).
A--10--X--0--B
| | |
| 10 |
| | |
+---0--Y--0--+
The above case would result in a path from A to Y to B to X, unless you increase the cost of the A and B edges (by 21).
A--31--X--21--B
| | |
| 10 |
| | |
+---21--Y--21-+
Now it goes from A to X to Y to B. Also make sure you remove any edges (A,B) (if they exist).
来源：https://stackoverflow.com/questions/24956968/implementation-of-a-particular-travelling-salesman-variation