Tipărirea celei mai lungi secvențe bitonice

Problema celei mai lungi subsecvențe bitonice este de a găsi cea mai lungă subsecvență a unei secvențe date astfel încât să crească mai întâi și apoi să descrească. O secvență sortată în ordine crescătoare este considerată Bitonică cu partea descrescătoare ca fiind goală. În mod similar, secvența de ordine descrescătoare este considerată Bitonică, cu partea crescătoare ca fiind goală. Exemple:

  Input:    [1 11 2 10 4 5 2 1]   Output:   [1 2 10 4 2 1] OR [1 11 10 5 2 1] OR [1 2 4 5 2 1]   Input:    [12 11 40 5 3 1]   Output:   [12 11 5 3 1] OR [12 40 5 3 1]   Input:    [80 60 30 40 20 10]   Output:   [80 60 30 20 10] OR [80 60 40 20 10] 

În anterior post pe care am discutat despre problema celei mai lungi subsecvențe bitonice. Cu toate acestea, postarea a acoperit doar codul legat de găsirea sumei maxime a subsecvenței crescătoare, dar nu și a construcției subsecvenței. În această postare vom discuta despre cum să construim cea mai lungă subsecvență bitonic în sine. Fie arr[0..n-1] tabloul de intrare. Definim vectorul LIS astfel încât LIS[i] este el însuși un vector care stochează cea mai lungă subsecvență crescătoare a arr[0..i] care se termină cu arr[i]. Prin urmare, pentru un index i LIS[i] poate fi scris recursiv ca -

LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] where   j  < i   and arr[j]  < arr[i] = arr[i] if there is no such j 

De asemenea, definim un vector LDS astfel încât LDS[i] este el însuși un vector care stochează cea mai lungă subsecvență descrescătoare a arr[i..n] care începe cu arr[i]. Prin urmare, pentru un index i LDS[i] poate fi scris recursiv ca -

LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} where   j > i   and arr[j]  < arr[i] = arr[i] if there is no such j 

De exemplu, pentru matrice [1 11 2 10 4 5 2 1]

LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1 
LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1 

Prin urmare, cea mai lungă subsecvență bitonică poate fi

LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1] 

Mai jos este implementarea ideii de mai sus - 

C++
   /* Dynamic Programming solution to print Longest    Bitonic Subsequence */   #include          using     namespace     std  ;   // Utility function to print Longest Bitonic   // Subsequence   void     print  (  vector   <  int  >&     arr       int     size  )   {      for  (  int     i     =     0  ;     i      <     size  ;     i  ++  )      cout      < <     arr  [  i  ]      < <     ' '  ;   }   // Function to construct and print Longest   // Bitonic Subsequence   void     printLBS  (  int     arr  []     int     n  )   {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      vector   <  vector   <  int  >>     LIS  (  n  );      // initialize LIS[0] to arr[0]      LIS  [  0  ].  push_back  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )      {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )      {      if     ((  arr  [  j  ]      <     arr  [  i  ])     &&      (  LIS  [  j  ].  size  ()     >     LIS  [  i  ].  size  ()))      LIS  [  i  ]     =     LIS  [  j  ];      }      LIS  [  i  ].  push_back  (  arr  [  i  ]);      }      /* LIS[i] now stores Maximum Increasing    Subsequence of arr[0..i] that ends with    arr[i] */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      vector   <  vector   <  int  >>     LDS  (  n  );      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  push_back  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )      {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )      {      if     ((  arr  [  j  ]      <     arr  [  i  ])     &&      (  LDS  [  j  ].  size  ()     >     LDS  [  i  ].  size  ()))      LDS  [  i  ]     =     LDS  [  j  ];      }      LDS  [  i  ].  push_back  (  arr  [  i  ]);      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      reverse  (  LDS  [  i  ].  begin  ()     LDS  [  i  ].  end  ());      /* LDS[i] now stores Maximum Decreasing Subsequence    of arr[i..n] that starts with arr[i] */      int     max     =     0  ;      int     maxIndex     =     -1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of LIS[i] + size      // of LDS[i] - 1      if     (  LIS  [  i  ].  size  ()     +     LDS  [  i  ].  size  ()     -     1     >     max  )      {      max     =     LIS  [  i  ].  size  ()     +     LDS  [  i  ].  size  ()     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]     LIS  [  maxIndex  ].  size  ()     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  size  ());   }   // Driver program   int     main  ()   {      int     arr  []     =     {     1       11       2       10       4       5       2       1     };      int     n     =     sizeof  (  arr  )     /     sizeof  (  arr  [  0  ]);      printLBS  (  arr       n  );      return     0  ;   }   
Java
   /* Dynamic Programming solution to print Longest    Bitonic Subsequence */   import     java.util.*  ;   class   GFG      {      // Utility function to print Longest Bitonic      // Subsequence      static     void     print  (  Vector   <  Integer  >     arr       int     size  )         {      for     (  int     i     =     0  ;     i      <     size  ;     i  ++  )      System  .  out  .  print  (  arr  .  elementAt  (  i  )     +     ' '  );      }      // Function to construct and print Longest      // Bitonic Subsequence      static     void     printLBS  (  int  []     arr       int     n  )         {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      @SuppressWarnings  (  'unchecked'  )      Vector   <  Integer  >[]     LIS     =     new     Vector  [  n  ]  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LIS  [  i  ]     =     new     Vector   <>  ();      // initialize LIS[0] to arr[0]      LIS  [  0  ]  .  add  (  arr  [  0  ]  );      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )         {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )         {      if     ((  arr  [  i  ]     >     arr  [  j  ]  )     &&         LIS  [  j  ]  .  size  ()     >     LIS  [  i  ]  .  size  ())         {      for     (  int     k     :     LIS  [  j  ]  )      if     (  !  LIS  [  i  ]  .  contains  (  k  ))      LIS  [  i  ]  .  add  (  k  );      }      }      LIS  [  i  ]  .  add  (  arr  [  i  ]  );      }      /*    * LIS[i] now stores Maximum Increasing Subsequence     * of arr[0..i] that ends with arr[i]    */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      @SuppressWarnings  (  'unchecked'  )      Vector   <  Integer  >[]     LDS     =     new     Vector  [  n  ]  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ]     =     new     Vector   <>  ();      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ]  .  add  (  arr  [  n     -     1  ]  );      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )         {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )         {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&         LDS  [  j  ]  .  size  ()     >     LDS  [  i  ]  .  size  ())      for     (  int     k     :     LDS  [  j  ]  )      if     (  !  LDS  [  i  ]  .  contains  (  k  ))      LDS  [  i  ]  .  add  (  k  );      }      LDS  [  i  ]  .  add  (  arr  [  i  ]  );      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      Collections  .  reverse  (  LDS  [  i  ]  );      /*    * LDS[i] now stores Maximum Decreasing Subsequence     * of arr[i..n] that starts with arr[i]    */      int     max     =     0  ;      int     maxIndex     =     -  1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of       // LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ]  .  size  ()     +     LDS  [  i  ]  .  size  ()     -     1     >     max  )      {      max     =     LIS  [  i  ]  .  size  ()     +     LDS  [  i  ]  .  size  ()     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]       LIS  [  maxIndex  ]  .  size  ()     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]       LDS  [  maxIndex  ]  .  size  ());      }      // Driver Code      public     static     void     main  (  String  []     args  )         {      int  []     arr     =     {     1       11       2       10       4       5       2       1     };      int     n     =     arr  .  length  ;      printLBS  (  arr       n  );      }   }   // This code is contributed by   // sanjeev2552   
Python3
   # Dynamic Programming solution to print Longest   # Bitonic Subsequence   def   _print  (  arr  :   list     size  :   int  ):   for   i   in   range  (  size  ):   print  (  arr  [  i  ]   end  =  ' '  )   # Function to construct and print Longest   # Bitonic Subsequence   def   printLBS  (  arr  :   list     n  :   int  ):   # LIS[i] stores the length of the longest   # increasing subsequence ending with arr[i]   LIS   =   [  0  ]   *   n   for   i   in   range  (  n  ):   LIS  [  i  ]   =   []   # initialize LIS[0] to arr[0]   LIS  [  0  ]  .  append  (  arr  [  0  ])   # Compute LIS values from left to right   for   i   in   range  (  1     n  ):   # for every j less than i   for   j   in   range  (  i  ):   if   ((  arr  [  j  ]    <   arr  [  i  ])   and   (  len  (  LIS  [  j  ])   >   len  (  LIS  [  i  ]))):   LIS  [  i  ]   =   LIS  [  j  ]  .  copy  ()   LIS  [  i  ]  .  append  (  arr  [  i  ])   # LIS[i] now stores Maximum Increasing   # Subsequence of arr[0..i] that ends with   # arr[i]   # LDS[i] stores the length of the longest   # decreasing subsequence starting with arr[i]   LDS   =   [  0  ]   *   n   for   i   in   range  (  n  ):   LDS  [  i  ]   =   []   # initialize LDS[n-1] to arr[n-1]   LDS  [  n   -   1  ]  .  append  (  arr  [  n   -   1  ])   # Compute LDS values from right to left   for   i   in   range  (  n   -   2     -  1     -  1  ):   # for every j greater than i   for   j   in   range  (  n   -   1     i     -  1  ):   if   ((  arr  [  j  ]    <   arr  [  i  ])   and   (  len  (  LDS  [  j  ])   >   len  (  LDS  [  i  ]))):   LDS  [  i  ]   =   LDS  [  j  ]  .  copy  ()   LDS  [  i  ]  .  append  (  arr  [  i  ])   # reverse as vector as we're inserting at end   for   i   in   range  (  n  ):   LDS  [  i  ]   =   list  (  reversed  (  LDS  [  i  ]))   # LDS[i] now stores Maximum Decreasing Subsequence   # of arr[i..n] that starts with arr[i]   max   =   0   maxIndex   =   -  1   for   i   in   range  (  n  ):   # Find maximum value of size of LIS[i] + size   # of LDS[i] - 1   if   (  len  (  LIS  [  i  ])   +   len  (  LDS  [  i  ])   -   1   >   max  ):   max   =   len  (  LIS  [  i  ])   +   len  (  LDS  [  i  ])   -   1   maxIndex   =   i   # print all but last element of LIS[maxIndex] vector   _print  (  LIS  [  maxIndex  ]   len  (  LIS  [  maxIndex  ])   -   1  )   # print all elements of LDS[maxIndex] vector   _print  (  LDS  [  maxIndex  ]   len  (  LDS  [  maxIndex  ]))   # Driver Code   if   __name__   ==   '__main__'  :   arr   =   [  1     11     2     10     4     5     2     1  ]   n   =   len  (  arr  )   printLBS  (  arr     n  )   # This code is contributed by   # sanjeev2552   
C#
   /* Dynamic Programming solution to print longest    Bitonic Subsequence */   using     System  ;   using     System.Linq  ;   using     System.Collections.Generic  ;   class     GFG      {      // Utility function to print longest Bitonic      // Subsequence      static     void     print  (  List   <  int  >     arr       int     size  )         {      for     (  int     i     =     0  ;     i      <     size  ;     i  ++  )      Console  .  Write  (  arr  [  i  ]     +     ' '  );      }      // Function to construct and print longest      // Bitonic Subsequence      static     void     printLBS  (  int  []     arr       int     n  )         {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      List   <  int  >  []     LIS     =     new     List   <  int  >  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LIS  [  i  ]     =     new     List   <  int  >  ();      // initialize LIS[0] to arr[0]      LIS  [  0  ].  Add  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )         {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )         {      if     ((  arr  [  i  ]     >     arr  [  j  ])     &&         LIS  [  j  ].  Count     >     LIS  [  i  ].  Count  )         {      foreach     (  int     k     in     LIS  [  j  ])      if     (  !  LIS  [  i  ].  Contains  (  k  ))      LIS  [  i  ].  Add  (  k  );      }      }      LIS  [  i  ].  Add  (  arr  [  i  ]);      }      /*    * LIS[i] now stores Maximum Increasing Subsequence     * of arr[0..i] that ends with arr[i]    */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      List   <  int  >  []     LDS     =     new     List   <  int  >  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ]     =     new     List   <  int  >  ();      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  Add  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )         {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )         {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&         LDS  [  j  ].  Count     >     LDS  [  i  ].  Count  )      foreach     (  int     k     in     LDS  [  j  ])      if     (  !  LDS  [  i  ].  Contains  (  k  ))      LDS  [  i  ].  Add  (  k  );      }      LDS  [  i  ].  Add  (  arr  [  i  ]);      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ].  Reverse  ();      /*    * LDS[i] now stores Maximum Decreasing Subsequence     * of arr[i..n] that starts with arr[i]    */      int     max     =     0  ;         int     maxIndex     =     -  1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of       // LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ].  Count     +     LDS  [  i  ].  Count     -     1     >     max  )      {      max     =     LIS  [  i  ].  Count     +     LDS  [  i  ].  Count     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]     LIS  [  maxIndex  ].  Count     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  Count  );      }      // Driver Code      public     static     void     Main  (  String  []     args  )         {      int  []     arr     =     {     1       11       2       10       4       5       2       1     };      int     n     =     arr  .  Length  ;      printLBS  (  arr       n  );      }   }   // This code is contributed by PrinciRaj1992   
JavaScript
   // Function to print the longest bitonic subsequence   function     _print  (  arr       size  )     {      for     (  let     i     =     0  ;     i   <  size  ;     i  ++  )     {      process  .  stdout  .  write  (  arr  [  i  ]  +  ' '  );      }   }   // Function to construct and print the longest bitonic subsequence   function     printLBS  (  arr       n  )     {      // LIS[i] stores the length of the longest increasing subsequence ending with arr[i]      let     LIS     =     new     Array  (  n  );      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LIS  [  i  ]     =     [];      }      // initialize LIS[0] to arr[0]      LIS  [  0  ].  push  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  let     i     =     1  ;     i      <     n  ;     i  ++  )     {      // for every j less than i      for     (  let     j     =     0  ;     j      <     i  ;     j  ++  )     {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&     LIS  [  j  ].  length     >     LIS  [  i  ].  length  )     {      LIS  [  i  ]     =     LIS  [  j  ].  slice  ();      }      }      LIS  [  i  ].  push  (  arr  [  i  ]);      }      // LIS[i] now stores the Maximum Increasing Subsequence of arr[0..i] that ends with arr[i]      // LDS[i] stores the length of the longest decreasing subsequence starting with arr[i]      let     LDS     =     new     Array  (  n  );      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LDS  [  i  ]     =     [];      }      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  push  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  let     i     =     n     -     2  ;     i     >=     0  ;     i  --  )     {      // for every j greater than i      for     (  let     j     =     n     -     1  ;     j     >     i  ;     j  --  )     {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&     LDS  [  j  ].  length     >     LDS  [  i  ].  length  )     {      LDS  [  i  ]     =     LDS  [  j  ].  slice  ();      }      }      LDS  [  i  ].  push  (  arr  [  i  ]);      }      // reverse the LDS vector as we're inserting at the end      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LDS  [  i  ].  reverse  ();      }      // LDS[i] now stores the Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i]      let     max     =     0  ;      let     maxIndex     =     -  1  ;      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      // Find maximum value of size of LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ].  length     +     LDS  [  i  ].  length     -     1     >     max  )     {      max     =     LIS  [  i  ].  length     +     LDS  [  i  ].  length     -     1  ;      maxIndex     =     i  ;      }      }      // print all but      // print all but last element of LIS[maxIndex] array      _print  (  LIS  [  maxIndex  ].  slice  (  0       -  1  )     LIS  [  maxIndex  ].  length     -     1  );      // print all elements of LDS[maxIndex] array      _print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  length  );   }   // Driver program   const     arr     =     [  1       11       2       10       4       5       2       1  ];   const     n     =     arr  .  length  ;   printLBS  (  arr       n  );   

Ieșire:

1 11 10 5 2 1 

Complexitatea timpului soluția de programare dinamică de mai sus este O(n 2 ). Spațiu auxiliar folosit de program este O(n 2 ).