Minimum capacity cut reduction from digraph with two edge weight sets Given a digraph $G$ and $f, g : E(G) \mapsto \mathbb{R}$, how would you find a cut $(X,\bar{X})$ with $s \in X$ and $t \in \bar{X}$ such that $\sum_{e \in \delta^+(X)}{f(e)} - \sum_{e \in \delta^-(X)}{g(e)}$ is minimized?

$\delta^+(X)$ is the set of edges going out of $X$ and $\delta^-(X)$ is the set of edges going into $X$.

I can solve for the case where $f = g$, simply by constructing a network flow graph with all edges adjacent to either $s$ or $t$. Since $\sum_{e \in \delta^+(X)}{f(e)} - \sum_{e \in \delta^-(X)}{f(e)}$ can be rewritten as $\sum_{e \in \delta^+(X)}{f(e)} + \sum_{e \notin \delta^-(X)}{f(e)} + constant$ then a graph can be constructed with an edge between $s$ and all $v \notin \{s,t\}$ with capacity equal to $\sum_{e \in \delta^-(v)}{f(e)}$  and an edge between all $v \notin \{s,t\}$ and $t$ with capacity equal to $\sum_{e \in \delta^+(v)}{f(e)}$. Then dealing with negative $f$ is simply a matter of adding the capacity to the $(v,t)$ edge instead of the $(s,v)$ edge and vice versa.

However this solution does not work when $f \neq g$ and I can't come up with an answer for this case. 

The solution works for the case that $f=g$ because if we have an edge $(a,b)$, in the case that $a \in X, b \in \bar{X}$ then the edge is going out of X and not going into X, the contribution to the cut capacity is $2f((a,b))$ as per the formula containing the constant above. Then if you add $b$ to $X$, $a$ no longer contributes to the sum over edges going out of $X$ and so $f((a,b))$ is removed and you are left with $f((a,b))$. If $f \neq g$, then you are instead left with $f((a,b)) − g((a,b))$ which is incorrect.

In terms of what I have tried so far for the general case, the case that $g(x) &lt; 0$ is trivial and the edge can simply be reversed and the capacity multiplied by $-1$ so that it is positive.

In the case that both $f,g &gt; 0$ then I consider this local portion of the graph:


Let $f = f(a,b)$ and $g = g(a,b)$ in the above graph.
I can then transform it to this network graph:


Where the capacities are given by


$(s,b) = f$
$(b,a) = f - g$
$(a,t) = f$


Clearly this is only applicable in the case where $f &gt; g$. Then in the case that $a$ but not $b$ is added to $X$ we have a net increase of $f$. When $b$ but not $a$ is added to $X$ we have a net decrease of $g$. When both $a$ and $b$ are added to $X$ there is no net change.

However I cannot reconcile this with the case where $f &lt; g$ or $f &lt; 0$.
This problem is NP-hard. 

Consider the $f=0$ case, then you are trying to minimize $-\Sigma_{e\in\delta^-(X)}g(e)$, i.e. maximize $\Sigma_{e\in\delta^-(X)}g(e)$. This is exactly the maximum cut problem, which is NP-hard.