It isn't an Hamiltonian cycle. That's when you must visit each vertex exactly once. This is "shortest path" which can be done pretty quick.
A question: do you have to go left to right or can you go right to left also? Put another way, do those edges just go left to right, or can they go right to left too?
If it's left to right only then it's easy to solve. Start by figuring the cheapest way to get from the 3rd column to the 4th. For each node in the 3rd column, record both the edge that's cheapest and the total cost of the path.
Now go to the 2nd column and compute the cheapest path. For each node in the 2nd column, look at the nodes in the 3rd column that it can reach and choose the path with the least total cost. The key here is that you don't have to look at the 4th column, only the 3rd. Again, record the TOTAL cost and the path. Note that for the path, you only have to record the edge to the 3rd column, not the whole path to the 4th column. To reconstruct the path, you just go to that node in the 3rd column and get the edge for it's
Now, do the same for the 1st column: get the cheapest path from each node in the 1st column.
Finally, compare the total cost of each node in the 1st column to find the one with the cheapest path.
This is the same as Dijkstra's algorithm with the small change that you know ahead of time which nodes you to examine when.