Gegeben sei eine quadratische Matrix der Größe N*N, in der jede Zelle mit einem bestimmten Preis verbunden ist. Ein Pfad ist als eine bestimmte Folge von Zellen definiert, die in der Zelle oben links beginnt, sich nur nach rechts oder unten bewegt und in der Zelle unten rechts endet. Wir wollen einen Pfad mit dem maximalen Durchschnitt aller vorhandenen Pfade finden. Der Durchschnitt wird als Gesamtkosten dividiert durch die Anzahl der auf dem Pfad besuchten Zellen berechnet. 

Beispiele:  

 Input : Matrix = [1 2 3   
  4 5 6   
  7 8 9]   
 Output : 5.8   
 Path with maximum average is 1 -> 4 -> 7 -> 8 -> 9   
 Sum of the path is 29 and average is 29/5 = 5.8   
 
 Eine interessante Beobachtung ist, dass nur Bewegungen nach unten und nach rechts zulässig sind. Wir benötigen N-1 Bewegungen nach unten und N-1 Bewegungen nach rechts, um das Ziel zu erreichen (ganz rechts unten). Jeder Pfad von der oberen linken Ecke zur unteren rechten Ecke erfordert also 2N – 1 Zellen. In    Durchschnitt    Wert ist der Nenner fest und wir müssen nur den Zähler maximieren. Daher müssen wir grundsätzlich den maximalen Summenpfad finden. Die Berechnung der maximalen Pfadsumme ist ein klassisches Problem der dynamischen Programmierung. Wenn dp[i][j] die maximale Summe bis zur Zelle (i j) von (0 0) darstellt, aktualisieren wir dp[i][j] wie folgt in jeder Zelle (i j).  
  
 for all i 1  <= i  <= N   
  dp[i][0] = dp[i-1][0] + cost[i][0];    
 for all j 1  <= j  <= N   
  dp[0][j] = dp[0][j-1] + cost[0][j];    
 otherwise    
  dp[i][j] = max(dp[i-1][j] dp[i][j-1]) + cost[i][j];    
 
 Sobald wir die maximale Summe aller Pfade erhalten haben, teilen wir diese Summe durch (2N – 1) und erhalten unseren maximalen Durchschnitt.   
  
    Durchführung:    
 C++   
   //C/C++ program to find maximum average cost path   #include          using     namespace     std  ;   // Maximum number of rows and/or columns   const     int     M     =     100  ;   // method returns maximum average of all path of   // cost matrix   double     maxAverageOfPath  (  int     cost  [  M  ][  M  ]     int     N  )   {      int     dp  [  N  +  1  ][  N  +  1  ];      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ];      /* Initialize first column of total cost(dp) array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -1  ][  0  ]     +     cost  [  i  ][  0  ];      /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -1  ]     +     cost  [  0  ][  j  ];      /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <=     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     max  (  dp  [  i  -1  ][  j  ]      dp  [  i  ][  j  -1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N  -1  ][  N  -1  ]     /     (  2  *  N  -1  );   }   /* Driver program to test above functions */   int     main  ()   {      int     cost  [  M  ][  M  ]     =     {     {  1       2       3  }      {  6       5       4  }      {  7       3       9  }      };      printf  (  '%f'       maxAverageOfPath  (  cost       3  ));      return     0  ;   }   
  Java   
   // JAVA Code for Path with maximum average   // value   import     java.io.*  ;   class   GFG     {          // method returns maximum average of all      // path of cost matrix      public     static     double     maxAverageOfPath  (  int     cost  [][]        int     N  )      {      int     dp  [][]     =     new     int  [  N  +  1  ][  N  +  1  ]  ;      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ]  ;          /* Initialize first column of total cost(dp)    array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -  1  ][  0  ]     +     cost  [  i  ][  0  ]  ;          /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -  1  ]     +     cost  [  0  ][  j  ]  ;          /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     Math  .  max  (  dp  [  i  -  1  ][  j  ]        dp  [  i  ][  j  -  1  ]  )     +     cost  [  i  ][  j  ]  ;          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N  -  1  ][  N  -  1  ]     /     (  2     *     N     -     1  );      }          /* Driver program to test above function */      public     static     void     main  (  String  []     args  )         {      int     cost  [][]     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }};          System  .  out  .  println  (  maxAverageOfPath  (  cost       3  ));      }   }   // This code is contributed by Arnav Kr. Mandal.   
  C#   
   // C# Code for Path with maximum average   // value   using     System  ;   class     GFG     {          // method returns maximum average of all      // path of cost matrix      public     static     double     maxAverageOfPath  (  int     []  cost        int     N  )      {      int     []  dp     =     new     int  [  N  +  1    N  +  1  ];      dp  [  0    0  ]     =     cost  [  0    0  ];          /* Initialize first column of total cost(dp)    array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i       0  ]     =     dp  [  i     -     1    0  ]     +     cost  [  i       0  ];          /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0       j  ]     =     dp  [  0    j     -     1  ]     +     cost  [  0       j  ];          /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i       j  ]     =     Math  .  Max  (  dp  [  i     -     1       j  ]      dp  [  i    j     -     1  ])     +     cost  [  i       j  ];          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N     -     1       N     -     1  ]     /     (  2     *     N     -     1  );      }          // Driver Code      public     static     void     Main  ()         {      int     []  cost     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }};          Console  .  Write  (  maxAverageOfPath  (  cost       3  ));      }   }   // This code is contributed by nitin mittal.   
  JavaScript   
    <  script  >      // JavaScript Code for Path with maximum average value          // method returns maximum average of all      // path of cost matrix      function     maxAverageOfPath  (  cost       N  )      {      let     dp     =     new     Array  (  N  +  1  );      for     (  let     i     =     0  ;     i      <     N     +     1  ;     i  ++  )      {      dp  [  i  ]     =     new     Array  (  N     +     1  );      for     (  let     j     =     0  ;     j      <     N     +     1  ;     j  ++  )      {      dp  [  i  ][  j  ]     =     0  ;      }      }      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ];          /* Initialize first column of total cost(dp)    array */      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -  1  ][  0  ]     +     cost  [  i  ][  0  ];          /* Initialize first row of dp array */      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -  1  ]     +     cost  [  0  ][  j  ];          /* Construct rest of the dp array */      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     Math  .  max  (  dp  [  i  -  1  ][  j  ]      dp  [  i  ][  j  -  1  ])     +     cost  [  i  ][  j  ];          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     dp  [  N  -  1  ][  N  -  1  ]     /     (  2     *     N     -     1  );      }          let     cost     =     [[  1       2       3  ]      [  6       5       4  ]      [  7       3       9  ]];          document  .  write  (  maxAverageOfPath  (  cost       3  ));    <  /script>   
  PHP   
      // Php program to find maximum average cost path    // method returns maximum average of all path of    // cost matrix    function   maxAverageOfPath  (  $cost     $N  )   {   $dp   =   array  (  array  ())   ;   $dp  [  0  ][  0  ]   =   $cost  [  0  ][  0  ];   /* Initialize first column of total cost(dp) array */   for   (  $i   =   1  ;   $i    <   $N  ;   $i  ++  )   $dp  [  $i  ][  0  ]   =   $dp  [  $i  -  1  ][  0  ]   +   $cost  [  $i  ][  0  ];   /* Initialize first row of dp array */   for   (  $j   =   1  ;   $j    <   $N  ;   $j  ++  )   $dp  [  0  ][  $j  ]   =   $dp  [  0  ][  $j  -  1  ]   +   $cost  [  0  ][  $j  ];   /* Construct rest of the dp array */   for   (  $i   =   1  ;   $i    <   $N  ;   $i  ++  )   {   for   (  $j   =   1  ;   $j    <=   $N  ;   $j  ++  )   $dp  [  $i  ][  $j  ]   =   max  (  $dp  [  $i  -  1  ][  $j  ]  $dp  [  $i  ][  $j  -  1  ])   +   $cost  [  $i  ][  $j  ];   }   // divide maximum sum by constant path    // length : (2N - 1) for getting average    return   $dp  [  $N  -  1  ][  $N  -  1  ]   /   (  2  *  $N  -  1  );   }   // Driver code   $cost   =   array  (  array  (  1     2     3  )   array  (   6     5     4  )   array  (  7     3     9  )   )   ;   echo   maxAverageOfPath  (  $cost     3  )   ;   // This code is contributed by Ryuga   ?>   
  Python3   
   # Python program to find    # maximum average cost path   # Maximum number of rows    # and/or columns   M   =   100   # method returns maximum average of    # all path of cost matrix   def   maxAverageOfPath  (  cost     N  ):   dp   =   [[  0   for   i   in   range  (  N   +   1  )]   for   j   in   range  (  N   +   1  )]   dp  [  0  ][  0  ]   =   cost  [  0  ][  0  ]   # Initialize first column of total cost(dp) array   for   i   in   range  (  1     N  ):   dp  [  i  ][  0  ]   =   dp  [  i   -   1  ][  0  ]   +   cost  [  i  ][  0  ]   # Initialize first row of dp array   for   j   in   range  (  1     N  ):   dp  [  0  ][  j  ]   =   dp  [  0  ][  j   -   1  ]   +   cost  [  0  ][  j  ]   # Construct rest of the dp array   for   i   in   range  (  1     N  ):   for   j   in   range  (  1     N  ):   dp  [  i  ][  j  ]   =   max  (  dp  [  i   -   1  ][  j  ]   dp  [  i  ][  j   -   1  ])   +   cost  [  i  ][  j  ]   # divide maximum sum by constant path   # length : (2N - 1) for getting average   return   dp  [  N   -   1  ][  N   -   1  ]   /   (  2   *   N   -   1  )   # Driver program to test above function   cost   =   [[  1     2     3  ]   [  6     5     4  ]   [  7     3     9  ]]   print  (  maxAverageOfPath  (  cost     3  ))   # This code is contributed by Soumen Ghosh.   
    
  Ausgabe   
5.200000  
 
    Zeitkomplexität    : AN   2   ) für die gegebene Eingabe N   
    Hilfsraum:    AN   2   ) für gegebene Eingabe N.  
  
    Methode - 2: Ohne zusätzlichen N*N-Speicherplatz zu verwenden     
  
 
 Wir können das Eingabekosten-Array als DP verwenden, um die Ans zu speichern. Auf diese Weise benötigen wir kein zusätzliches DP-Array und keinen zusätzlichen Speicherplatz.  
  
 
 Eine Beobachtung ist, dass die einzigen zulässigen Bewegungen nach unten und nach rechts sind. Wir benötigen N-1 Abwärtsbewegungen und N-1 Rechtsbewegungen, um das Ziel (ganz rechts unten) zu erreichen. Jeder Pfad von der oberen linken Ecke zur unteren rechten Ecke erfordert also 2N – 1 Zelle. In    Durchschnitt    Wert ist der Nenner fest und wir müssen nur den Zähler maximieren. Daher müssen wir grundsätzlich den maximalen Summenpfad finden. Die Berechnung der maximalen Pfadsumme ist ein klassisches Problem der dynamischen Programmierung. Außerdem benötigen wir nach der Berechnung von dp[i][j] keinen vorherigen Kosten[i][j]-Wert, sodass wir den Kosten[i][j]-Wert so ändern können, dass wir keinen zusätzlichen Platz für dp[i][j] benötigen.  
  
 for all i 1  <= i  < N   
  cost[i][0] = cost[i-1][0] + cost[i][0];    
 for all j 1  <= j  < N   
  cost[0][j] = cost[0][j-1] + cost[0][j];    
 otherwise    
  cost[i][j] = max(cost[i-1][j] cost[i][j-1]) + cost[i][j];    
 
 Nachfolgend finden Sie die Umsetzung des oben genannten Ansatzes:  
 C++   
   // C++ program to find maximum average cost path   #include          using     namespace     std  ;   // Method returns maximum average of all path of cost matrix   double     maxAverageOfPath  (  vector   <  vector   <  int  >>  cost  )   {      int     N     =     cost  .  size  ();      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ];      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ];      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <=     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     max  (  cost  [  i     -     1  ][  j  ]     cost  [  i  ][  j     -     1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1  ][  N     -     1  ]     /     (  2     *     N     -     1  );   }   // Driver program   int     main  ()   {      vector   <  vector   <  int  >>     cost     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }      };      cout      < <     maxAverageOfPath  (  cost  );      return     0  ;   }   
  Java   
   // Java program to find maximum average cost path   import     java.io.*  ;   class   GFG     {      // Method returns maximum average of all path of cost      // matrix      static     double     maxAverageOfPath  (  int  [][]     cost  )      {      int     N     =     cost  .  length  ;      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ]  ;      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ]  ;      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     Math  .  max  (  cost  [  i     -     1  ][  j  ]        cost  [  i  ][  j     -     1  ]  )      +     cost  [  i  ][  j  ]  ;      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1  ][  N     -     1  ]     /     (  2     *     N     -     1  );      }      // Driver program      public     static     void     main  (  String  []     args  )      {      int  [][]     cost      =     {     {     1       2       3     }     {     6       5       4     }     {     7       3       9     }     };      System  .  out  .  println  (  maxAverageOfPath  (  cost  ));      }   }   // This code is contributed by karandeep1234   
  C#   
   // C# program to find maximum average cost path   using     System  ;   class     GFG     {      // Method returns maximum average of all path of cost      // matrix      static     double     maxAverageOfPath  (  int  [     ]     cost  )      {      int     N     =     cost  .  GetLength  (  0  );      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i       0  ]     =     cost  [  i       0  ]     +     cost  [  i     -     1       0  ];      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0       j  ]     =     cost  [  0       j     -     1  ]     +     cost  [  0       j  ];      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  i       j  ]     =     Math  .  Max  (  cost  [  i     -     1       j  ]      cost  [  i       j     -     1  ])      +     cost  [  i       j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1       N     -     1  ]     /     (  2     *     N     -     1  );      }      // Driver program      static     void     Main  (  string  []     args  )      {      int  [     ]     cost      =     {     {     1       2       3     }     {     6       5       4     }     {     7       3       9     }     };      Console  .  WriteLine  (  maxAverageOfPath  (  cost  ));      }   }   // This code is contributed by karandeep1234   
  JavaScript   
   // Method returns maximum average of all path of cost matrix   function     maxAverageOfPath  (  cost  )   {      let     N     =     cost  .  length  ;      // Initialize first column of total cost array      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ];      // Initialize first row of array      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ];      // Construct rest of the array      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  let     j     =     1  ;     j      <=     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     Math  .  max  (  cost  [  i     -     1  ][  j  ]     cost  [  i  ][  j     -     1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  cost  [  N     -     1  ][  N     -     1  ])     /     (  2.0     *     N     -     1  );   }   // Driver program   let     cost     =     [[  1       2       3  ]      [  6       5       4  ]      [  7       3       9  ]];   console  .  log  (  maxAverageOfPath  (  cost  ))   // This code is contributed by karandeep1234.   
  Python3   
   # Python program to find maximum average cost path   from   typing   import   List   def   maxAverageOfPath  (  cost  :   List  [  List  [  int  ]])   ->   float  :   N   =   len  (  cost  )   # Initialize first column of total cost array   for   i   in   range  (  1     N  ):   cost  [  i  ][  0  ]   =   cost  [  i  ][  0  ]   +   cost  [  i   -   1  ][  0  ]   # Initialize first row of array   for   j   in   range  (  1     N  ):   cost  [  0  ][  j  ]   =   cost  [  0  ][  j   -   1  ]   +   cost  [  0  ][  j  ]   # Construct rest of the array   for   i   in   range  (  1     N  ):   for   j   in   range  (  1     N  ):   cost  [  i  ][  j  ]   =   max  (  cost  [  i   -   1  ][  j  ]   cost  [  i  ][  j   -   1  ])   +   cost  [  i  ][  j  ]   # divide maximum sum by constant path   # length : (2N - 1) for getting average   return   cost  [  N   -   1  ][  N   -   1  ]   /   (  2   *   N   -   1  )   # Driver program   def   main  ():   cost   =   [[  1     2     3  ]   [  6     5     4  ]   [  7     3     9  ]]   print  (  maxAverageOfPath  (  cost  ))   if   __name__   ==   '__main__'  :   main  ()   
    
  Ausgabe   
5.2  
 
    Zeitkomplexität:    O(N*N)   
    Hilfsraum:    O(1)