Transpose Rows to column
I have inserted your values into a table variable just to so I could pivot from it later.

The code came out like this:

SET TRANSACTION ISOLATION LEVEL READ UNCOMMITTED
SET NOCOUNT ON
SET DEADLOCK_PRIORITY LOW

declare @t table(srv nvarchar(128),dates date,[count] int)

insert into @t (srv,dates,[count])
select 'SRV1',cast ('03/25/2018' as date) ,1   union all
select 'SRV1',cast ('04/06/2018' as date) ,59  union all
select 'SRV1',cast ('04/07/2018' as date) , 9  union all
select 'SRV1',cast ('04/13/2018' as date) , 1  union all
select 'SRV1',cast ('04/15/2018' as date) , 1  union all
select 'SRV1',cast ('04/20/2018' as date) , 4  union all
select 'SRV1',cast ('04/21/2018' as date) , 2  union all
select 'SRV1',cast ('04/27/2018' as date) , 5  union all
select 'SRV1',cast ('04/28/2018' as date) , 5  union all
select 'SRV1',cast ('04/29/2018' as date) , 1  union all
select 'SRV1',cast ('04/30/2018' as date) ,62  



select *
from (

       select srv  
              ,dates
              ,[count]
          from @t

) radhe
pivot (sum([count]) for dates in (
  [2018-04-01], 
  [2018-04-02], 
  [2018-04-03], 
  [2018-04-04], 
  [2018-04-05], 
  [2018-04-06], 
  [2018-04-07], 
  [2018-04-08], 
  [2018-04-09], 
  [2018-04-10], 
  [2018-04-11], 
  [2018-04-12], 
  [2018-04-13], 
  [2018-04-14], 
  [2018-04-15],
  [2018-04-16], 
  [2018-04-17], 
  [2018-04-18], 
  [2018-04-19], 
  [2018-04-20], 
  [2018-04-21], 
  [2018-04-22], 
  [2018-04-23], 
  [2018-04-24], 
  [2018-04-25], 
  [2018-04-26], 
  [2018-04-27], 
  [2018-04-28], 
  [2018-04-29], 
  [2018-04-30])) as krishna
where 1=1


there are things that could be changed, but that is all up to you,
plus if you want to get rid of the nulls and replace them by zeros
please have a look at the following question and the answers at:

How to replace (null) values with 0 output in PIVOT

And that will also give you some ideas as how to produce all this in a dynamic way.