Su tarea es crear la cadena objetivo dada. Comenzando con una cadena que está vacía, tendrá que agregarle caracteres, hasta que su cadena sea la misma que queremos. Puede agregar un carácter al final de su cadena con costo x, o puede clonar su cadena con costo y. Lo que queremos es la forma más barata de hacer esto.

Casos de prueba

targetString , appendcost, clonecost -> totalcost

"bb", 1, 2 -> 2
"bbbb", 2, 3 -> 7
"xzxpcxzxpy", 10, 11 -> 71
"abababab", 3, 5 -> 16
"abababab", 3, 11 -> 23

Casco , 25 bytes

φ?ö▼z+:⁴∞²m⁰§:h§δf`€otṫḣ0

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Las entradas están en el orden agregar costo, costo de clonación, objetivo.

Explicación

φ?ö▼z+:⁴∞²m⁰§:h§δf`€otṫḣ0  Two explicit inputs and one implicit.
                           Example: 2, 3, s="abab"
φ                          Make a recursive function and call it on s:
 ?                      0   If s is empty, return 0.
  ö▼z+:⁴∞²m⁰§:h§δf`€otṫḣ    Otherwise do this.
                       ḣ    Prefixes: ["a","ab","aba","abab"]
                    otṫ     Suffixes except the first one: ["bab","ab","b"]
               §δf`€        Keep those prefixes that have the corresponding suffix as substring: ["ab","aba"]
            §:h             Prepend s minus last character: ["aba","ab","aba"]
          m⁰                Recurse on each: x=[6,4,6]
        ∞²                  Repeat the clone cost: [3,3,3,..
      :⁴                    Prepend append cost: [2,3,3,3,..
    z+                      Add component-wise to x: [8,7,9]
   ▼                        Minimum: 7

Python 2 , 112 bytes

f=lambda s,a,c,i=0:i<len(s)and min([a+f(s,a,c,i+1)]+[c+f(s,a,c,n)for n in range(i+1,len(s)+1)if s[i:n]in s[:i]])

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JavaScript (ES6), 123111 bytes

Toma entrada como (x)(y)(s).

x=>y=>m=g=([s,...r],o='',c=0)=>s?[...r,g(r,o+s,c+x)].map(_=>s+=r.shift(~o.search(s)&&g(r,o+s,c+y)))|m:m=m<c?m:c

Pruébalo en línea!

Comentado

x => y =>                    // x = 'append' cost; y = 'clone' cost
m =                          // m = minimum cost, initialized to a non-numeric value
                             //     this is what will eventually be returned
g = (                        // g = recursive function taking:
  [s,                        //   - the input string split into s = next character
      ...r],                 //     and r[] = array of remaining characters
  o = '',                    //   - o = output string
  c = 0                      //   - c = current cost
) =>                         //
  s ?                        // if s is defined:
    [ ...r,                  //   split a copy of r
      g(r, o + s, c + x)     //   do a recursive call with an 'append' operation
    ].map(_ =>               //   iterate as many times as there are remaining characters
                             //   in r[], + 1
      s +=                   //     append to s
        r.shift(             //     the next character picked from the beginning of r[]
          ~o.search(s) &&    //     if s is found in o,
          g(r, o + s, c + y) //     do a recursive call with a 'clone' operation
        )                    //     (note that both s and r are updated *after* the call)
    ) | m                    //   end of map(); return m
  :                          // else:
    m = m < c ? m : c        //   update m to min(m, c)

R , 192 185 bytes

f=function(s,a,c,k=0,x="",R=substring,N=nchar,p=R(s,1,1:N(s)))'if'(!N(s),k,{y={};for(i in c(R(s,1,1),p[mapply(grepl,p,x)]))y=min(y,f(R(s,N(i)+1),a,c,k+'if'(any(y),c,a),paste0(x,i)));y})

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Código desenrollado y explicación:

# s is the current remaining string (at the beginning is equal to the target string)
# a is the append cost
# c is the clone cost
# k is the current cost (at the beginning is zero)
# x is the partially constructed string (at the beginning is empty)
f=function(s,a,c,k=0,x=""){
  # store in p all the possible prefixes of s
  p = substring(s,1,1:nchar(s))
  # if s is empty return the current cost k
  if(!nchar(s))
    k
  else{
    y={}
    # prepend the first letter of s (=append operation) to  
    # the prefixes in p that are contained in x (=clone operations)
    for(i in c(substring(s,1,1),p[mapply(grepl,p,x)])){
      # perform first the append then the clone operations and recurse, 
      # storing the cost in y if lower than previous
      # (if y is NULL is an append operation otherwise is a clone, we use the right costs)
      y = min(y,f(substring(s,nchar(i)+1),a,c,k+'if'(any(y),c,a),paste0(x,i)))
    }
    # return the current cost
    y
  }
}