Topographic Strings

J, 87 79 72 70 67 57 56 characters

'( ) 'charsub|.|:(+/\@('('&=-')'&=)(],~' '$~[)"0])1!:1[1

Takes input from the keyboard. Example:

   '( ) 'charsub|.|:(+/\@('('&=-')'&=)(],~' '$~[)"0])1!:1[1
((1 2)(3 (4 5) moo)) (i (lik(cherries)e (woohoo)))
          4 5                cherries    woohoo
  1 2  3       moo       lik          e
                      i

Explanation:

This explanation is based on the first version of my program:

|.|:('( ) 'charsub x)((' '$~{.@]),[{~{:@])"1(('('&([:+/=)-')'&([:+/=))\,.i.@#)x=.1!:1[1

x=.1!:1[1 take input from keyboard and put it into x for later

(('('&([:+/=)-')'&([:+/=))\,.i.@#) creates a list of all indeces into the string (i.@#) and stitches (,.) it together with the result of the (('('&([:+/=)-')'&([:+/=))\ verb.

(('('&([:+/=)-')'&([:+/=))\ this verb is applied to all the prefixes of the string (so on input hello it would apply to h,he,hel,hell, and hello. It is a fork, which counts the number of open brackets ('('&([:+/=) and then subtracts the number of close brackets ')'&([:+/=). This gives me list of indeces into the string and the level the character at that index should be at in the output. On simple input this gives me the following:

   (('('&([:+/=)-')'&([:+/=))\,.i.@#)x=.1!:1[1
(one(two(three)))
1  0
1  1
1  2
1  3
2  4
2  5
2  6
2  7
3  8
3  9
3 10
3 11
3 12
3 13
2 14
1 15
0 16

((' '$~{.@]),[{~{:@])"1 this is a verb which takes the list I just generated and also the output of ('( ) 'charsub x) (which just does a string replacement to replace all brackets with spaces in x). It takes the tail of each item of the list {:@] and uses it as an index into the string to get the character [{~{:@]. Then it prefixes it , with the number of spaces as indicated by the head of each item in the list(' '$~{.@]). On the previous example this gives me:

   ('( ) 'charsub x)((' '$~{.@]),[{~{:@])"1(('('&([:+/=)-')'&([:+/=))\,.i.@#)x=.1!:1[1
(one(two(three)))

 o
 n
 e

  t
  w
  o

   t
   h
   r
   e
   e

I then transpose the array |: and reverse it |. to get the desired output.

GolfScript 69

0:§;{.'()'?))3%(.§+:§' ':s*\@s\if\n}%n/.{,}%$)\;:μ;{.,μ\-s*\+}%zip n*

Online demo here.

Explanation:

0:§;                # declare the variable §, representing the 
                    # current vertical level and initialize it at 0

{                   # iterate for each char in the string:

    .'()'?))3% (    # add on the stack the amount by which
                    # the current vertical level should be 
                    # adjusted:
                    #   * +1 if the character is '('
                    #   * -1 if the character is ')'
                    #   * 0 otherwise

    .§+:§           # adjust the value of §

    ' ':s*          # add as many spaces as § tells us
                    # and save the space in variable s

    \@s\if\         # return current char, if it's printable,
                    # or a space if it's '(' or ')'

    n               # add a newline char

}%

n/                  # split by newline char; now we have 
                    # an array of strings on the stack.
                    # Each string is a vertical line of the
                    # final output.

.{,}%$)\;:μ;        # Iterate through the strings and find the
                    # maximum length

{
    .,μ\-s*\+       # Add spaces at the end to make all the strings 
                    # the same length
}%

zip                 # Transpose the strings

n*                  # Join the transposed strings by newline characters

APL (59)

⊖↑{⊃,/T\¨⍨⍵×P=0}¨R∘=¨(⍴T)∘⍴¨⍳⌈/R←1++\P←+/¨1 ¯1∘ר'()'∘=¨T←⍞

I have assumed that the 'base' needs to be usable as well. (i.e. (a(b))c(d) is valid). If this is not necessary two characters can be saved.

Explanation:

• T←⍞: store a line of input in T
• '()'∘=¨T: for each character in T, see if it is an opening or closing parenthesis. This gives a list of lists of booleans.
• 1 ¯1∘ר: multiply the second element in each of these lists by -1 (so an opening parenthesis is 1, a closing one is -1 and any other character is 0).
• +/¨: take the sum of each inner list. We now have the ∆y value for each character.
• P←: store in P.
• R←1++\P: take a running total of P, giving the height for each character. Add one to each character so that characters outside of the parentheses are on the first line.
• (⍴T)∘⍴¨⍳⌈/R: for each possible y-value, make a list as long as T, consisting of only that value. (i.e. 1111..., 2222...., etc.)
• R∘=¨: for each element in this list, see if it is equal to R. (For each level, we now have a list of zeroes and ones corresponding to whether or not a character should appear on that level).
• ⍵×P=0: for each of these lists, set it to zero if P is not zero at that spot. This gets rid of the characters with a non-zero delta-y, so that gets rid of the parentheses.
• ⊃,/T\¨⍨: for each depth, select from T the characters that should appear.
• ⊖↑: create a matrix and put it right-side-up.