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The Wiki article doesn't specify, but that code will only solve the 'loopy' version of k-shortest-paths, where paths are not required to be simple.

The simple path version of the problem is harder: you'll want to look at something like Yen's algorithm, which does clever filtering to avoid repeated points when generating paths. Yen's algorithm can use Dijkstra's algorithm as a subroutine, but any other shortest-path algorithm can also be used instead.

There is no obvious way to modify Dijkstra's algorithm to solve the k-shortest-simple-paths problem. You'd need to track the paths in the priority queue (which is already done in your posted code), but there's an exponential upper bound on the number of times each vertex can be explored.

Here, if count[u] <= K puts an upper bound of K+1 on the number of times a vertex can be explored, which works for the non-simple path case. On the other hand, a direct modification of Dijkstra's algorithm for simple paths would, in the worst case, require you to explore a node once for each of the 2^(V-1) possibilities of which nodes had been previously visited (or possibly a slightly smaller exponential).

Here's a great answer, thanks to @NicoSchertler:

"You can push prev and next to travQueue even if they are nullptr. When you reach a nullptr in your iteration, add the dummy value to the result and two more nullptr to travQueue for the non-existing children."

And here's my code for it:

You could solve max flow with +1 sources at x and y, and -1 sinks at u and v.

Remove the edge e, and set all other edges to capacity 1.

A simple path from x to y via edge e exists if and only if you can find a flow of 2 in this new max flow problem.

Split each vertex v[i] in the original graph into two vertices, a[i] and b[i].

For each undirected edge between v[i] and v[j] in the original, add directed edges b[j] to a[i] and b[i] to a[j] with capacity 1.

Also add a directed edge from a[i] to b[i] with capacity 1 for each vertex v[i].

The idea is that flow must always arrive at an a[i] vertex, and leave from a b[i] vertex, after passing through the capacity 1 bottleneck from a[i] to b[i]. This ensures that each vertex can only be used once.

With this new graph, proceed as for the edge-independent case.

Based on what you said I understand your question as follows: How the to handle Nodes that on the edges like (x=0,y=0), (x=9,y=5) or (x=9.y=9) ..... you should handle 8 cases

for case of left top corner the node only have 2 neighbors so set the top and left neighbors Null

First of all, I would say, my solution below is a brute-force method, thus only working well for graphs of small size, i.e., not many vertices or arcs.

If you have large networks, you should refer to some more advanced algorithms, e.g., https://link.springer.com/article/10.1007/s40747-018-0079-7

Since you have 6 arcs and 5 vertices, you only need to remove 2 arcs out of 6 to find the spanning tree. There would be combn(6,2) options, and you can delete those edge combinations one by one to check if a spanning tree remains

I understand that bidirectional BFS consists of, given start and goal nodes, alternately expanding layers of nodes from start and goal, until a node in the middle has been reached from both ends. The shortest path from start to goal is then understood to be the shortest from start to middle node, continued by the shortest from middle to goal. I can see that less nodes may need to be expanded as compared with a standard BFS approach. However,

• It is easier to implement standard BFS (sBFS) than to implement a bidirectional BFS (bBFS for short). Simple is often good, as it is easier to code and to later verify its correctness.
• If the graph is directed and unweighted, sBFS guarantees that it will find the shortest path from start to goal in minimal steps; bBFS is not guaranteed to work with directed graphs.
• After running sBFS, you can reconstruct shortest paths from the start to all nodes that are at most 1 step before the goal node (that is, before the search was stopped). This may be valuable in and of itself. Running bBFS does not generate such a list.

With this in mind, I would argue the bBFS is only useful for a very narrow case (where, depending on the graph, it is expected to perform better than sBFS), and sBFS is both simpler and useful in a larger range of scenarios.

Can You share that code which take O(n^2) ?

Anyway, This approach only take O(n), Which is better then O(n log n)!

You could do something like this:

You might need margin-bottom instead of padding-bottom;

By using padding you don't seperate them, padding works kind of from-inside.

You may read about box-model here to understand this.