Changing the problem of the Chinese postman_

To prepare well for my spring exams now, I study and experiment with graph issues.

I’m already used to typical problems, such as "Traveling Salesman", but when I delved into the "problem of the Chinese postman" and its variations, I immediately felt that an important aspect of the problem was missing: the aspect of availability is limited capacity, so you need to go back to the office station After a certain number of letters has been successfully delivered (to get new ones). What about finding the shortest path?

I was very intrigued by CPP because of its relevance and ease of use for real life, but I think adding this aspect will make it even more real practice.

I would be grateful for any help in finding the shortest path in an undirected graph that visits each edge at least once (CPP), since it has to return to the starting point (post station) after a certain number of letters.

EDIT (description of the original CPP): “The problem of the Chinese postman or the problem of checking the route is to find the shortest closed path or circuit that visits each edge of the (connected) undirected graph. When the graph has an Euler circuit (closed walk that covers each edge once), this scheme is the optimal solution. If the graph is not Euler, it must contain vertices of odd degree. By the acknowledgment lemma there must be an even number of these vertices. To solve the postman problem, we first find the smallest T-connection. We make the graph Euler by doubling the T-union. The solution to the postman problem in the original graph is obtained by finding the Eulerian scheme for the new graph. " Src: wikipedia.org