The Skyline Problem‍​​

I'm just starting to learn J, so here goes my first attempt at golf:

103 62 49
46 characters

   b =: 8 3 $ 1 11 5 2 6 7 3 13 9 12 7 16 14 3 25 19 18 22 23 13 29 24 4 28
   ,(,.{&s)I.s~:_1|.s=:0,~>./(1&{*{.<:[:i.{:)"1 b
1 11 3 13 9 0 12 7 16 3 19 18 22 3 23 13 29 0

although I'm sure someone who actually knows the language well could shorten this by quite a bit

An explanation of the code:

   NB. list numbers up to right bound of the building
   ([: i. {:) 14 3 25  
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
   NB. compare to left bound of building and then multiply by height
   (1&{ * {. <: [: i. {:) 14 3 25 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 3 3 3 3 3 3 3
   NB. apply to each row of b, note how shorter entries are padded with 0s
   (1&{ * {. <: [: i. {:)"1 b
0 11 11 11 11  0  0  0  0 0 0 0 0 0 0 0 0 0 0  0  0  0 0  0  0  0  0  0  0
0  0  6  6  6  6  6  0  0 0 0 0 0 0 0 0 0 0 0  0  0  0 0  0  0  0  0  0  0
...
   NB. collapse to find max, add a 0 to the end for rotate later, assign to s
   ] s =: 0 ,~ >./ (1&{ * {. <: [: i. {:)"1 b
0 11 11 13 13 13 13 13 13 0 0 0 7 7 7 7 3 3 3 18 18 18 3 13 13 13 13 13 13 0
   NB. rotate s right 1 and then compare to s to find where height changes
   s ~: _1 |. s
0 1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1
   NB. find indices of all differences
   I. s ~: _1 |. s
1 3 9 12 16 19 22 23 29
   NB. pair each index with the height of the building there
   (,. {&s) I. s ~: _1 |. s
 1 11
 3 13
 9  0
...
   NB. and finally flatten the list
   , (,. {&s) I. s ~: _1 |. s
1 11 3 13 9 0 12 7 16 3 19 18 22 3 23 13 29 0

Python, 89 characters, also using Triptych's 5001 cheat:

B=[[1,11,5],[2,6,7],[3,13,9],[12,7,16],[14,3,25],[19,18,22],[23,13,29],[24,4,28]]

x=o=0
while x<5001:
 n=max([H for L,H,R in B if L<=x<R]+[0])
 if n-o:print x,n,
 o=n;x+=1

Replacing 5001 by max(map(max,B))+1 to allow (almost) arbitrarily large cities leaves 102 characters.

Revision History:

• saved two chars as described in John Pirie's comment
• saved one char as MahlerFive suggested

Python: 115 characters

Like the OP, I'm not including the declaration of the data, but I am counting whitespace.

D = [(1,11,5), (2,6,7), (3,13,9), (12,7,16), 
 (14,3,25), (19,18,22), (23,13,29), (24,4,28)]

P=[max([0]+[h for s,h,e in D if s<=x<e])for x in range(5001)]
for i,x in enumerate(P[1:]):
   if x!=P[i]:print i+1,x,

Note that I am using the link provided by the OP as the exact definition of the problem. For instance, I cheat a bit by assuming there is not building coordinate over 5000, and that all coordinates are positive integers. The original post is not tightly constrained enough for this to be fun, in my opinion.

edit: thanks to John Pirie for the tip about collapsing the list construction into the printing for loop. How'd I miss that?!

edit: I changed range(1+max(zip(*D)[2])) to range(5001) after deciding the use the exact definition given in the original problem. The first version would handle buildings of arbitrary positive integers (assuming it all fit into memory).

edit: Realized I was overcomplicating things. Check my revisions.

BTW - I have a hunch there's a much more elegant, and possibly shorter, way to do this. Somebody beat me!