Algorithm for solving incremental max flow problem
You can do it in $\small \mathcal{O}(m + n)$ time where $\small m$ and $\small n$ are the # of edges and vertices respectively. Let the edge to be updated be $\small e = (u, v)$. 


If you increment the capacity of $\small e$ by $\small 1$, the maximum flow increments by at most $\small 1$. Hence, starting with the current max flow $\small f$, you only need to find an augmenting path in the residual graph one more time, which costs $\small \mathcal{O}(m + n)$ time. Recall that each augmentation increments the flow value by at least $\small 1$. Therefore, it is guaranteed after the augmentation (if any), the flow attained is maximum.
If you decrement the capacity of $\small e$ by $\small 1$, the maximum flow decrements by at most $\small 1$. If the new capacity (after decreased by $\small 1$) $\small c(e) \geq f(e)$, where $\small f(e)$ is the current flow value of $\small e$, then nothing need to be conducted. Otherwise, we have to decrease the flow value of $\small e$. Let $\small s \rightarrow u$ be a simple path from $\small s$ to $\small u$, in which all edges have positive flow values and similarly define $\small v \rightarrow t$. We then decrease the flow values by $\small 1$ for all edges in the path $\small s \rightarrow u \rightarrow v \rightarrow t$. Finally, we again search for an augmenting path in the resultant residual graph.