Algorithm 3 Primal-dual heuristic for finding an HDSF
1:	Initialization
2:	$F_k\leftarrow \varnothing$, ${\mathcal{C}}_k\leftarrow \{\left\{v\right\}:v\in V_k\}$ for $k=1,2$
3:	All the vertices are unmarked.
4:	All the dual variables are set to zero.
5:	$activ{e}_k\left(\left\{v\right\}\right)\leftarrow 1$, $\forall v\in V_k$, for $k=1,2$
6:	$activ{e}_k\left(\left\{d_k\right\}\right)\leftarrow 0$, for $k=1,2$
7:	Main loop
8:	while there exists any active component in ${\mathcal{C}}_1,{\mathcal{C}}_2$ do
9:	for k = 1, 2 do
10:	Find an edge $e_k=(i,j)\in E_k$ with $i\in C_i,j\in C_j$, where $C_i,C_j\in {\mathcal{C}}_k,C_i\ne C_j$ that minimizes ${\epsilon }_k=\frac{W_k^{+}cos{t}_{ij}^k-dua{l}_k\left(j\right)}{activ{e}_k\left(C_j\right)}$
11:	end for
12:	Let the corresponding $C_j\in {\mathcal{C}}_k$ be $S_k$, while $S=\{S_1,S_2\}$ satisfies $S_1\supseteq S_2$ and $S_1,S_2$ are active.
13:	$F_k\leftarrow \left\{e_k\right\}\cup F_k$
14:	Increase the dual variables of $S_k$ by ${\epsilon }_k$.
15:	if$e_k$ forms a new strongly connected component and the component is not reachable from $d_k$ then
16:	Let the strongly connected component be an active component.
17:	else if $e_k$ makes any vertex $v\in S_k$ reachable from $d_k$ then
18:	Let the depot and all vertices that are reachable from the depot be an inactive component.
19:	if $e_k=e_1$ then
20:	Deactivate all subsets of this component in ${\mathcal{C}}_2$.
21:	else
22:	Mark all vertices in the supersets of this component in ${\mathcal{C}}_1$. Deactivate it if the corresponding component consists of all marked vertices.
23:	end if
24:	else
25:	Deactivate $S_k$.
26:	end if
27:	if there exists no $S=\{S_1,S_2\}$ that can be chosen that satisfies the given conditions and there exists any inactive set without an incoming edge that is not connected to the depot then
28:	Pick an inactive component for each k that consists of marked vertices that have incoming or outgoing edges. Combine those connected components until the new component does not have any incoming edges.
29:	end if
30:	end while
31:	Pruning
32:	$P_k\leftarrow \mathrm{all}\mathrm{vertices}\mathrm{that}\mathrm{are}\mathrm{only}\mathrm{reachable}\mathrm{from}{d}_k\mathrm{for}k=1,2$
33:	$P_c\leftarrow \mathrm{the}\mathrm{vertices}\mathrm{that}\mathrm{are}\mathrm{reachable}\mathrm{from}\mathrm{both}{d}_1\mathrm{and}{d}_2$
34:	Let $T_k$ be the minimum directed spanning tree of $P_k\mathrm{for}k=1,2$.
35:	Let $C\left(T_k\right)$ be the sum of the edge costs present in $T_k$.
36:	while$P_c$ is not empty do
37:	Find the shortest edge $e_k$ that makes a vertex in $P_c$ reachable from the vertices in $P_k$ for each k.
38:	if $C\left(T_1\right)\le C\left(T_2\right)$ then
39:	Add $e_1$ to $T_1$, remove the corresponding vertex from $P_c$, and add it to $P_1$.
40:	else
41:	Add $e_2$ to $T_2$, remove the corresponding vertex from $P_c$, and add it to $P_2$.
42:	end if
43:	end while