For this problem there are two sets. Set $C$ consists of incidents where each incident $c_i$ has a location and time of occurence. Set $V$ is a set of vehicles where each vehicle $v_j$ has a location. For every pair of ($v_j$, $c_i$) the driving time from vehicle to incident is known. So the input is an undirected complete bipartite graph $G=(V \cup C,E)$ with the cost of the edges being the duration to go from $v_j \rightarrow c_i \rightarrow v_j$.
The problem to solve is as follows: Every incident has to be connected to some vehicle such that the total cost (driving duration to all incidents) will be minimized. Obviously a vehicle can only be dispatched to two incidents if he is back from one incident before the second incidents has occurred.
Because this is an offline problem, we know when all the incidents of set $C$ will occure. Due to this we can pre-calculate for each vehicle $v_j$ and each incident $c_i$, which other incidents are unable to connect to vehicle $v_j$ when vehicle $v_j$ is connected to incident $c_i$. So for every pair $(v_j,c_i)$ there is a set of incidents $I_{c_i,v_j}$ that cannot be connected to vehicle $v_j$ if incident $c_i$ is already connected to vehicle $v_j$.
An assumption is that there are always enough vehicles so that every incident can be connected to some vehicle and that waiting for availability of vehicles is not allowed.
I think this problem can be solved by using a min-cost flow algorithm. I just don't see how to formulate it yet. We can first look at just one vehicle since all the vehicles are indepent from eachother. $\textbf{So if we would model the min-cost flow for one vehicle, how would it look like?}$
To get some feeling for the problem, the input is as follows (edge costs are not in the image): complete bipartite graph
Now we add the source $s$ and target $t$, and only look at one vehicle. To illustrate the problem, see the following picture. In the picture there are lines between incidents if they cannot be connected to the same vehicle. Due to this, in the picture there are 2 sets that share no vehicles (namely the first three incidents and the last two incidents). Because there is no overlap between the sets it works to draw an edge to each set with cost zero and capacity one. Let the balance of every incident be -1. Then only one incident of each set can be matched in such a construction. Here not every incident will be connected but that is because not all vehicles are in the picture. See figure here: example where min-cost flow works
Here we see that it has minimal cost to connect the vehicle to the second and fifth incident. Now imagine that the second incident is also connected to the fourth incident. Then it would still be feasible to connect the vehicle to the second and fifth incident, but this would not be feasible according to the flow construction in the picture.
So how to model this min-cost flow that the flow construction is correct for every input?