Problem Notes:

This is a very difficult, but fascinating, graph problem.  It might be a
stopper, and at best I can't see more than 3-5 teams getting it.

At first glance, it appears to be a very straightforward Network Flow problem
(and the sample input/output don't contradict this).  But, as it turns out,
it's not that simple!  The key advantage you have is that you can wait to
destroy tunnels until the spy has made certain moves.  If he has 5 paths to
choose from, but each one involves moving through a room with only 2 tunnels
coming off it (a chokepoint), only two tunnels ever need to be destroyed.

So, how to solve it?  The solution is still Network-Flow-ish, but more
involved.  Let us store a "d-number" for each room that will eventually become
the number of tunnels that would have to be destroyed if the spy started in
that room.

The d-numbers start at infinity.  They are iteratively lowered according to
this rule: from any starting room, if there exist x,y >= 0 such that x tunnels
can be destroyed to force all paths to the outside to pass through a room with
d-number <= y, then reduce the d-number to x+y (if it's higher).  Keep going
until applying this rule no longer lowers any d-numbers.

The proof that the algorithm works is below; I'm including it to prove the
correctness of my judge solution.  It's fairly non-trivial.  However, the
algorithm itself is fairly intuitive if you experiment with the problem for a
bit.  I think teams may come up with it and use it without actually proving
that it works.  (And, of course, there might be a simpler approach I haven't
seen...)

Note that applying d-number reductions is a simple application of network flow:
choose a y, set all rooms <= y to block flow, then calculate the min cut/max
flow to the outside of the lair, giving you x.  The one easy optimization I use
in my judge solution is that, for a given starting room, I don't clear the flow
information as I lower y.  The already-calculated flows remain valid as rooms
become unblocked.


PROOF OF ALGORITHM:

I'll prove that my algorithm works: in other words, that the final d-number of
each room is the minimum number of tunnels that ever need to be collapsed if
the spy starts in that room.

First, note that the algorithm that calculates the d-numbers is monotonic: the
d-numbers only ever decrease, and the order in which you apply the reductions
does not matter, because when a reduction is possible it never STOPS being
possible.  So the order in which reductions are applied can be completely
arbitrary, and the final d-numbers are always the same.

Now, one direction of the proof is simple: the d-number is clearly an upper
bound on the minimum number of tunnels we need to collapse.  We can stop the
spy from escaping from a room by applying the reduction that gave it its
d-number: collapse the chosen x tunnels, forcing the spy to pass through a room
with d-number <= y before escaping.  Since you still have y collapses left,
when he moves into one of those rooms, you can collapse its set of tunnels; and
so on.  He can't escape.

The tricky part of the proof is to show that it is NECESSARY to collapse this
many tunnels.  In particular, when using the above strategy, does collapsing a
tunnel in an early room allow us to collapse fewer tunnels later on?  I will
show that the answer is no.


LEMMA: Given graphs G and G', where G' is G with one tunnel "t" deleted, the
final d-numbers calculated by the above algorithm for G' are each either the
same as in G, or one less.

PROOF: All reductions in G apply in G' as well (perhaps even with one fewer
tunnel deleted), so clearly the final d-numbers in G' are no larger than in G.
Without loss of generality we can assume the algorithm first reduces all the
d-numbers in G' to the final values in G.  Suppose, for the sake of
contradiction, that a reduction eventually reduces a d-number in G' to at least
two lower than its final value in G.  Consider the first such reduction.

Let us examine one of the G'-specific reduction steps that came before this.
We know it reduced a d-number by one from that in G.  Let the reduction's room
be r, and consider its x,y values.  There was a way to delete x tunnels from G'
such that all escape paths from r pass through a room that had d-number <= y.
Now, since this reduction did NOT apply in G, one of the following must hold:

1) One of the escape paths that exists in G' with its x tunnels collapsed,
when considered in G, does NOT pass through a room with d-number <= y.
However, it must pass through a room which has a d-number of y+1 (since G and
G' d-numbers don't currently differ by more than one).  That room has already
been given a G'-specific reduction - and note that r's new G'-specific
d-number, x+y, is NO SMALLER than that room's already-reduced G'-specific
d-number, y.

2) ALL of the escape paths that exist in G' with its x tunnels collapsed, when
considered in G, pass through a room with a d-number <= y.  Then, of course,
x+1 tunnels can be collapsed in G to make an x+1,y reduction.  Now, there MUST
be a path in G from r to one of the endpoints "s" of t, a path that never
passes through any room with a d-number that is <= y.  Otherwise, since no
escape path could use the extra tunnel without passing through a room with
d-number <= y, the original x,y reduction would work in G - an impossibility.
Because of this path, the SAME x+1,y reduction works on room s as well: after
deleting the x+1 tunnels, there can't be any escape paths from s that never hit
a room with d-number <= y (or they could be extended to r as well).  So one of
the endpoints of t has a d-number in G that is <= x+y+1 - that is, AT MOST ONE
GREATER than r's new G'-specific d-number, x+y.

These two cases, taken together, show that there is an endpoint of t whose
d-number in G is no more than one greater than ANY of the G'-specific d-numbers
reduced so far.

Now, let us consider the new reduction, on room r with values x,y, that reduces
a d-number by at least two from that in G.  Once again there are two cases:

1) One of the escape paths that exists in G' with its x tunnels collapsed,
when considered in G, does NOT pass through a room with d-number <= y.  As
above, it must pass through a room which has a d-number of y+1 that was reduced
to y in G'.  But this means that one endpoint of t must have a d-number <= y+1
in G.  Thus, deleting the same x tunnels in G produces an x,y+1 reduction: all
escape paths that DON'T use t must pass through a room with d-number <= y+1
(the room with d-number <= y in the G' reduction) and all escape paths that DO
use t pass through one as well.  This is a contradiction, since the new
reduction is supposed to be 2 less than any possible in G.

2) ALL of the escape paths that exist in G' with its x tunnels collapsed, when
considered in G, pass through a room with a d-number <= y.  Then, as above,
an x+1,y reduction works.  Again, this is a contradiction.

So no such G'-specific reduction can exist, and the lemma is proved.


Now, suppose the spy starts in a room with final d-number n, and we only have
n-1 collapses available.  An escape strategy for the spy is as follows: we know
a path exists to the outside that passes through only rooms with d-number
>= n (otherwise the current room would have a 0,n-1 reduction).  The spy just
follows that route until we decide to collapse a tunnel.  Now, according to the
lemma, we can consider him to be starting in a new lair from a room with a
final d-number >= n-1 - and we now only have n-2 collapses available.  This
continues until we have 0 collapses available; he'll still be in a room with
d-number >= 1, from which he can escape.

Thus, the d-number of a room is exactly the number of collapses necessary to
trap a spy starting in that room.  QED.