Proving that Hamiltonian Cycle is reducible to a travelling problem? I chanced upon the following question online:


  A company has two trucks, and must deliver a number of parcels to a
  number of addresses. They want both drivers to be home at the end of
  the day. This gives the following decision problem.

Instance: Set V of locations, with for each pair of locations
v, w ∈ V , a distance d(v, w) ∈ N, a starting location s ∈ V ,
and an integer K.

  
  Question: Are there two cycles, that both start in s, such that every
  location in V is on at least one of the two cycles, and both cycles
  have length at most K?


The solution provided uses Hamiltonian Cycle to prove the parcel problem is NP-Complete.


  Given any instance of Hamiltonian Cycle with n vertices, construct the
  following special instance of the above problem. 
  
  
  Copy the graph and make the distance d(v, w) equal to 2 if there is    an edge in the graph, and 4, otherwise.
  Label one of these vertices as s. Add one vertex 0 such that d(s, 0)    = d(0, s) = n and d(v, 0) = d(0, v) = 2n + 1 for all v != s. The
  threshold K is equal to 2n.
  


The explanation provided is as follows:


  The only thing we have to do is to keep the other driver busy, which
  is easily taken care of by including an additional vertex with
  distance K/2 from s and distance K + 1 to all other addresses.


I tried constructing a simple graph of N = 3 whereby each vertex is connected by an edge of distance 2. Then I introduced a new node which is distance 4 away from the source vertex(i.e. starting point), and distance 7 from every other vertex. 

But I am unable to understand how this construction fits into the original decisional problem. What exactly does the solution mean by "keeping the other driver busy?"

Solution link.
Basically, Hamiltonian Cycle is the parcel problem with just one driver. To reduce Hamiltonian Cycle to the parcel problem with two drivers, you get one of the drivers to solve your Hamiltonian Cycle problem, and you make enough work for the other driver that they can't "cheat" by helping the first driver (but not so much work that the first driver has to help the second one).