Shortest sum of several paths - c++

Question

Let (x,p1,p2,...pn) denote that there is a directed edge from x to p1 (but not from p1 to x), a directed edge from x to p2, etc... and a directed edge from x to pn.
Suppose we have:
(a,b,c)
(b,d,c)
(c,d,e)
(f,e,h)
(g,f,h,i)
(i,j)
For simplicity, let's assume that the distances between all connected nodes are 1. Now there is no connection from a to g or vice versa, but we can have two people leave from a and g respectively and meet at a common point. We want to find the shortest sum of two paths that meet at a node such that one path begins at a and the other begins at g. A simple drawing of this directed graph will reveal that the answer are these two paths:
a->c->e
g->f->e
with a total distance of 4. How to write an algorithm that accepts a directed graph (as in the above) and two nodes (e.g. a and g in this case) and output this answer, which I suppose could be in the form std::make_pair({a,c,e}, {g,f,e})? I tried to adapt Dijkstra's Algorithm to do so but with no luck. I welcomes all ideas.
Edit: Above was a simplified version of the real problem. Here is the real problem, which I was reluctant to describe because an example graph would be too difficult to read. So I will describe it without an example graph.
Two points A and B are selected in a directed graph similar to the one above but larger. There is no connection from one point to the other, and neither is there a common point reachable from both A and B. However, we do know that there exist points (how many is not given) N1, N2, ... Nk such that there is a common point reachable from both A and N1, a common point reachable from both N1 and N2, ..., and a common point reachable from both Nk and B. We want to find these k+1 paths such that the total sum of these paths is minimum. That's the real problem.

Answer 1

You didn't adapt Dijkstra correctly. And you don't have to find dist(a,x) and dist(g,x) for every node x in every case.
In Dijkstra's algorithm, every node is considered Visited or Unvisited, and the search proceeds until the destination is Visited (or no further searching is possible).
In the variant, every node is Unvisited, Visited-By-A, Visited-By-B, or Visited-By-Both. When a node becomes Visited-By-Both, the sum of the paths to it is a limit on the search; the code will keep track of the smallest sum yet found, and terminate the search when the shortest path still being explored is longer than this sum.
I believe this search is O(V logV) in the worst case.
EDIT: The "real" problem.
We have A and B, and we are searching for {N}, {x} that minimizes
(|A, x1| + |N1, x1|) + (|N1, x2| + |N2, x2|) + (|N2, x3| + |N3, x3|)+ ... + (|Nk, xk| + |Nk, B|)
where |x,y| is the length of the shortest path from x to y.
Now consider a new graph, made by adding reverse edges to G: for every edge x->y in G, we add y->x (with the same weight, but all weights are 1 for our purposes) BUT we do not add backward edges leading to A. Then we REMOVE forward edges from B. Now find the shortest path on this new graph from A to B.
This path begins with a forward edge from A, ends with a backward edge to B, and is the shortest such path. Along the path, there must be nodes which the path enters on a forward edge and leaves along a backward edge; we label these xi. And likewise, there must be nodes which the path enters on a backward edge and leaves along a forward edge; we label these Ni. (There must be at least one N, because x1 cannot be xk, because we assume that there is no point forward-reachable from both A and B.)
If we break the path into all-forward and all-backward legs, we get
A-->x1, x1<--N1, N1-->x2, x2<--N2, N2-->x3, ..., xk<--Nk, Nk-->B
The length of this path is
|A,x1| + |N1, x1| + |N1,x2| + |N2,x2| + |N2,x3| + ... + |Nk,xk| + |Nk,B|, which is minimal for this choice of A,B.
Therefore these are the paths we're looking for (and they can be found by a simple O(V) transformation of the graph and a Dijkstra search (O(VlogV))).