"The sum of weight must not exceeds the pre given value M."

This is NP-hard by reduction from knapsack / subset sum. Think of a subset sum problem with numbers x1, ..., xn and you create n gadgets where to travel from vertex i to i+1 you either go through an edge of weight xi or an edge of weight 0. If the graph is undirected you can still enforce only traversing each edge once by setting a limit on the total number of edges you can travel (for example, each edge's weight is 10^67 + x, and the limit M is 5 * 10^67 + desired weight). What is the constraint on M?

1. If $$$M$$$ is small, we can use dp to solve it in $$$O(N^2*M)$$$. Define the state $$$dp[i][j]$$$ to indicate whether it is feasible to reach vertex $$$i$$$ with a distance cost of $$$j$$$. Whenever a state changes from infeasible to feasible, add this state to a queue. Each time a state $$$dp[u][j]$$$ is taken from the queue, use it to update all points $$$v$$$ connected to point $$$u$$$ (in a complete graph, that is, all vertices). Then dp[v][j + w[u][v]] ||= dp[u][j]. Obviously, there are $$$N*M$$$ states, and each state updates $$$N$$$ states, so the time complexity is $$$O(N^2*M)$$$. I'm not sure if there is a better way to do it; if there is, please let me know.
2. If $$$M$$$ is large, the problem might be NP-hard. If we consider each edge as a type of product, and the ability to traverse edges repeatedly is similar to the ability to choose an unlimited number of products, then it resembles the complete knapsack problem. I think the original problem is no simpler than this one, so it should also be NP-hard. Of course, I cannot prove this; it's just my guess.

This problem can be solved in polynomial time by Eppstein's algorithm.

A Leftist Heap gives the K shortest paths from start to end. Afterwards, it is simply a matter of continuously popping the path lengths from this heap (since it is least costliest at the root) until you reach the most expensive one under your budget.

See https://mirror.codeforces.com/blog/entry/102085 for additional details.

Note that the one modification you will have to make is blacklisting edges that go from anything --> source since you say we cannot go back to source.