How to find a cut in a graph with additional constraints?
The problem can be solved in polynomial time.  Here is one algorithm:

For each $s \in C$ and each $t \in V \setminus C$ such that $s \neq t$:

Find the minimum-cost $(s,t)$-cut.   By the max-flow min-cut theorem, this can be done in polynomial time using any maximum flow algorithm.

If the cost of this cut is &lt; 2, output it and halt.



For each $s_0 \in C$ and each $s_1 \in V \setminus C$ and each $t \in V \setminus \{s_0,s_1\}$:

Find the minimum-cost cut that contains both $s_0$ and $s_1$ in the left part and $t$ in the right part.  This can be done in polynomial time by modifying the graph to merge the two vertices $s_0,s_1$ into a new vertex $s'$, then finding a minimum $(s',t)$-cut in the modified graph.

If the cost of this cut is &lt; 2, output it (replacing $s'$ with $s_0,s_1$) and halt.



If you reach this point without halting, output that no such cut exists.


There may be more efficient algorithms by modifying algorithms for min-cut instead of min $(s,t)$-cut.  I don't know.  However, this suffices to show that the problem can be solved in polynomial time.

Proof of correctness: Suppose a cut $(S,V \setminus S)$ of the desired form exists.  Then there are only two ways we can have $S \ne C$:

Case 1: $C \setminus S \ne \emptyset$: In this case, pick any vertex in $C \cap S$ and call it $s^*$ (this can be done since $C \cap S \ne \emptyset$) and pick any vertex in $C \setminus S$ and call it $t^*$ (this can be done since $C \setminus S \ne \emptyset$).  Consider the minimum-cost $(s^*,t^*)$-cut, $(S^*,T^*)$.  By construction, $S^*$ has at least one element of $C$ (since $s^* \in S^*$) and $S^* \ne C$ (since $t^* \notin S^*$) and $\delta(S^*) \le \delta(S) &lt; 2$, so $(S^*,T^*)$ is a valid solution to your problem.  Moreover, it will be found by one of the iterations of the first for-loop, namely, when $s=s^*$ and $t=t^*$, so the algorithm will correctly find a solution.

Case 2: $S \setminus C \ne \emptyset$: In this case pick any vertex in $S \setminus C$ and call it $s^*_1$ (this can be done since $S \setminus C \ne \emptyset$), pick any vertex in $S \cap C$ and call it $s^*_0$ (this can be done since $S \cap C \ne \emptyset$), and pick any vertex in $V \setminus S$ and call it $t^*$ (this can be done since $S \ne V$).  Consider the minimum-cost cut $(S^*,T^*)$ such that $s^*_0 \in S^*$, $s^*_1 \in S^*$, and $t^* \in T^*$.  By construction, $S^*$ has at least one element of $C$ (since $s^*_0 \in S^*$) and $S^* \ne C$ (since $s^*_1 \in S^*$) and $\delta(S^*) \le \delta(S) &lt; 2$, so $(S^*,T^*)$ is a valid solution to your problem.  Moreover, it will be found by one of the iterations of the second for-loop, namely, when $s_0=s^*_0$ and $s_1=s^*_1$ and $t=t^*$, so the algorithm will correctly find a solution.


We see that in either case, if a solution exists, the algorithm will successfully output a valid solution; and those are the only two cases that can occur if a solution exists.
Conversely, if no valid solution exists, it is easy to see that the algorithm will correctly output that no cut exists.