JavaScript (Node.js), 38 bytes

a=>a.map(v=>(n+=v>p&&v-p,p=v),p=n=0)|n

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Simply a greedy algorithm which scan from left to right, only draw lines if needed, and draw it as long as possible.

Thanks Arnauld, save 2 3 bytes

05AB1E,  8 7  5 bytes

Saved 2 bytes thanks to @Adnan

0š¥þO

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How?

This is using the algorithm that was first found by @tsh. If you like this answer, make sure to upvote their answer as well!

Each time a skyscraper is lower than or as high as the previous one, it can be painted 'for free' by simply extending the brushstrokes.

For instance, painting skyscrapers $B$ and $C$ in the figure below costs nothing.

On the other hand, we need 2 new brushstrokes to paint skyscraper $E$, no matter if they're going to be reused after that or not.

For the first skyscraper, we always need as many brushstrokes as there are floors in it.

Turning this into maths:

$$S=h_0+sum_{i=1}^n max(h_i-h_{i-1},0)$$

If we prepend $0$ to the list, this can be simplified to:

$$S=sum_{i=1}^n max(h_i-h_{i-1},0)$$

Commented

0š¥þO     # expects a list of non-negative integers  e.g. [10, 9, 8, 9]

0š        # prepend 0 to the list                    -->  [0, 10, 9, 8, 9]

  ¥       # compute deltas                           -->  [10, -1, -1, 1]

   þ      # keep only values made of decimal digits

          # (i.e. without a minus sign)              -->  ["10", "1"]

    O     # sum                                      -->  11

Python 3, 37 bytes

lambda a:sum(a)-sum(map(min,a[1:],a))

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_{-5 bytes by switching to Python 3, thanks to Sarien}

Python 2, 47 43 42 bytes

lambda a:sum(a)-sum(map(min,a[1:],a[:-1]))

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Alt:

lambda a:sum(a)-sum(map(min,zip(a[1:],a)))

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Haskell, 32 bytes

(0%)

p%(h:t)=max(h-p)0+h%t

p%_=0

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An improvement on Lynn's solution that tracks the previous element p instead of looking at the next element. This makes the base case and recursive call shorter in exchange for needing to invoke (0%).

max(h-p)0 could be max h p-p for the same length.

Haskell, 35 bytes

f(a:b:c)=max(a-b)0+f(b:c)

f a=sum a

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Line 2 could be f[a]=a if I didn't also have to handle the case .

K (oK), 12 7 bytes

-5 bytes thanks to ngn!

A k (oK) port of Arnauld's 05AB1E solution (and tsh's JavaScript solution):

+/0|-':

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J, 15 bytes

A J port of Arnauld's 05AB1E solution (and tsh's JavaScript solution):

1#.0>./2-~/,]

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My naive solution:

J, 27 bytes

1#.2(1 0-:]),@|:@,~1$~"0]

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Haskell, 34 32 bytes

2 bytes trimmed by Lynn

g x=sum$max 0<$>zipWith(-)x(0:x)

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So to start we have zipWith(-). This takes two lists and produces a new list of their pairwise differences. We then combine it with x and (0:x). (0:) is a function that adds zero to the front of a list and by combining it with zipWith(-) we get the differences between consecutive elements of that list with a zero at the front. Then we turn all the negative ones to zero with (max 0<$>). This creates a new list where each element is the number of new strokes that has to be started at each tower. To get the total we just sum these with sum.

Japt, 8 bytes

-2 bytes from @Shaggy

mîT Õ¸¸è

Explanation

mîT Õ¸¸è      Full program. Implicit input U

                e.g. U = [2,0,2]

mîT             Map each item X and repeat T(0) X times

                     U = ["00","","00"]

    Õ           Transpose rows with columns

                     U = ["0 0","0 0"]

     ¸¸         Join using space and then split in space

                     U = ["0","0","0","0"]

        è       Return the count of the truthy values

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MATL, 8 bytes

0whd3Y%s

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Pretty much @Arnauld's algorithm. Saved one byte (thanks @LuisMendo) by casting to uint64 rather than selecting ) positive entries.

Japt, 7 6 bytes

änT fq

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1 byte saved thanks to Oliver.

änT xwT    :Implicit input of integer array

än         :Consecutive differences / Deltas

  T        :  After first prepending 0

    f      :Filter elements by

     q     :  Square root (The square root of a negative being NaN)

           :Implicitly reduce by addition and output

R, 30 bytes

sum(pmax(0,diff(c(0,scan()))))

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Porting of @Arnauld solution which in turn derives from @tsh great solution

Retina 0.8.2, 21 bytes

d+

$*

(1+)(?=,1)



1

Try it online! Link includes test cases. Explanation:

d+

$*

Convert to unary.

(1+)(?=,1)

Delete all overlaps with the next tower, which don't need a new stroke.

1

Count the remaining strokes.

Jelly, 5 bytes

A port of my 05AB1E answer, which itself is similar to @tsh JS answer.

ŻI0»S

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Commented

ŻI0»S    - main link, expecting a list of non-negative integers  e.g. [10, 9, 8, 9]

Ż        - prepend 0                                             -->  [0, 10, 9, 8, 9]

 I       - compute the deltas                                    -->  [10, -1, -1, 1]

  0»     - compute max(0, v) for each term v                     -->  [10, 0, 0, 1]

    S    - sum                                                   -->  11

Common Lisp, 88 87 bytes

(lambda(s)(let((o 0))(dolist(c s)(incf o(max 0 c))(mapl(lambda(y)(decf(car y)c))s))o))

non-minified

(lambda (skyline)

  (let ((output 0))

    (dolist (current-skyscraper-height skyline)

      (incf output (max 0 current-skyscraper-height))

      (mapl (lambda (skyscraper)

              (decf (car skyscraper) current-skyscraper-height))

            skyline))

    output)))

Test it

When one tower is painted, it takes a number of brushstrokes equal to it's height. These brushstrokes translate to all the following ones, which is indicated here by subtracting the current tower's height from all other towers (and itself, but that doesn't matter). If a following tower is shorter, then it will be pushed to a negative number, and this negative number will subsequently subtracted from the towers that follow (indicating brushstrokes that could not be translated from a previous tower to the next ones). It actually just subtracts the number from all tower heights, including previous ones, but this doesn't matter because we don't look at the previous ones again.

05AB1E, 13 10 bytes

Z>Lε@γPO}O

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Explanation:

Z            # Get the maximum of the (implicit) input-list

 >           # Increase it by 1 (if the list only contains 0s)

  L          # Create a list in the range [1, max]

   ε         # Map each value to:

    @        #  Check if this value is >= for each value in the (implicit) input

     γ       #  Split into chunks of adjacent equal digits

      P      #  Take the product of each inner list

       O     #  Take the sum

        }O   # And after the map: take the sum (which is output implicitly)

C# (Visual C# Interactive Compiler) with flag /u:System.Math, 47 bytes

n=>n.Select((a,i)=>i<1?a:Max(a-n[i-1],0)).Sum()

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Old version, with flag /u:System.Math, 63 bytes

n=>n.Aggregate((0,0),(a,b)=>(a.Item1+Max(0,b-a.Item2),b)).Item1

I feel like this solution is more elegant than the first one. It goes through the array with a two-value tuple as a starting value, picking up values, and stores the value before it in the second part of the tuple.

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Pyth, 8 bytes

s>#0.++0

Yet another port of @tsh's marvellous answer. Takes the sum (s) of the positive values (>#0) of the deltas (.+) of the input with 0 prepended (+0Q, trailing Q inferred).

Try it online here, or verify all the test cases at once here.

String joining method, 10 bytes

This was the solution I wrote before browsing the other answers.

lcj.t+d*LN

Test suite.

lcj.t+d*LNQ   Implicit: Q=eval(input()), b=<newline>, N=<quote mark>

              Trailing Q inferred

        L Q   Map each element of Q...

       * N    ... to N repeated that many times

     +b       Prepend a newline

   .t         Transpose, padding with spaces

  j           Join on newlines

 c            Split on whitespace

l             Take the length, implicit print

Clojure, 50 bytes

#((reduce(fn[[s n]i][(+(max(- i n)0)s)i])[0 0]%)0)

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#( ; begin anonymous function

    (reduce

        (fn [[s n] i] ; internal anonymous reducing function, destructures accumulator argument into a sum and the previous item

            [(+ (max (- i n) 0) s ; the sum part of the accumulator becomes the previous sum plus the larger of zero and the difference between the current number and the last one, which is how many new strokes need to be started at this point

            i]) ; ...and the previous item part becomes the current item

        [0 0] ; the initial value of the accumulator gives no strokes yet, and nothing for them to cover yet

        %) ; reduce over the argument to the function

    0) ; and get the sum element of the last value of the accumulator.

Java (JDK), 57 bytes

a->{int s=0,p=0;for(int x:a)s-=(x>p?p:x)-(p=x);return s;}

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Another port of tsh's Javascript answer. So make sure you've upvoted them.

Note that I used subtraction instead of addition because it allowed me to write (p=x) as right operand in the same statement.

MATL, 15 14 13 bytes

ts:<~"@Y'x]vs

Input is a column vector, using ; as separator.

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Explanation

t       % Implicit input: column vector. Duplicate

s       % Sum

:       % Range from 1 to that. Gives a row vector

<~      % Greater or equal? Element-wise with broadcast

"       % For each column

  @     %   Push current columnn

  Y'    %   Run-length encoding. Gives vector of values (0, 1) and vector of lengths

  x     %   Delete vector of lengths

]       % End

v       % Vertically concatenate. May give an empty array

s       % Sum. Implicit display

Perl 5, 21 bytes

$+=$_>$'&&$_-$';//}{

TIO

How

• 
  -p + }{ + $ trick


• 
  // matches empty string so that for next line postmatch $' will contain previous line


• 
  $+=$_>$'&&$_-$' to accumulate difference between current line and previous if current is greater than previous, (could also be written $+=$_-$' if$_>$', but perl doesn't parse $+=$_-$'if$_>$' the same)

Stax, 8 bytes

Φ┐Γ╟φφ.╨

Run and debug it

Uses the widely used approach from tsh's JavaScript solution.

Wolfram Language (Mathematica), 34 bytes

A port of @Arnauld's solution.

Tr@Select[{##,0}-{0,##},#>0&]&@@#&

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