This algorithm runs natively in O(V * E^2).
The description states that
The running time of O(V * E^2) is found by showing that each augmenting path can be found in O(E) time, that every time at least one of the E edges becomes saturated (an edge which has the maximum possible flow), [...]
My graph has the property that at there are exactly |V| edges with a limited capacity, that is, all remaining |E|-|V| edges have an unlimited capacity; any amount of flow can pass through them.
Given that those edges with an unlimited capacity can never become saturated, can I safely assume that a maximum of |V| augmenting paths will be found, and therefore the complexity reduced to O(V * E)?
If so, is there an adequat proof for the assumption made in the Wiki article? It only states There is an accessible proof in Introduction to Algorithms.