KEUZE VAN GEBIED

Beschouw een spel waarin je twee soorten krachten A en B hebt en er zijn 3 soorten Gebieden X Y en Z. Elke seconde moet je wisselen tussen deze gebieden en elk gebied heeft specifieke eigenschappen waardoor je kracht A en kracht B toenemen of afnemen. We moeten gebieden zo blijven kiezen dat onze overlevingstijd wordt gemaximaliseerd. De overlevingstijd eindigt wanneer een van de machten A of B minder dan 0 bereikt.

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Voorbeelden:

Initial value of Power A = 20 Initial value of Power B = 8 Area X (3 2) : If you step into Area X A increases by 3 B increases by 2 Area Y (-5 -10) : If you step into Area Y A decreases by 5 B decreases by 10 Area Z (-20 5) : If you step into Area Z A decreases by 20 B increases by 5 It is possible to choose any area in our first step. We can survive at max 5 unit of time by following these choice of areas : X -> Z -> X -> Y -> X

Dit probleem kan worden opgelost door recursie te gebruiken na elke tijdseenheid die we naar een bepaald gebied kunnen gaan, maar we zullen dat gebied kiezen, wat uiteindelijk leidt tot maximale overlevingstijd. Omdat recursie kan leiden tot het vaak oplossen van hetzelfde deelprobleem, zullen we het resultaat onthouden op basis van macht A en B. Als we hetzelfde paar machten A en B bereiken, zullen we het niet opnieuw oplossen, maar zullen we het eerder berekende resultaat nemen.

Hieronder vindt u de eenvoudige implementatie van bovenstaande aanpak.

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CPP

// C++ code to get maximum survival time #include    using namespace std; // structure to represent an area struct area {  // increment or decrement in A and B  int a b;  area(int a int b) : a(a) b(b)  {} }; // Utility method to get maximum of 3 integers int max(int a int b int c) {  return max(a max(b c)); } // Utility method to get maximum survival time int maxSurvival(int A int B area X area Y area Z  int last map<pair<int int> int>& memo) {  // if any of A or B is less than 0 return 0  if (A <= 0 || B <= 0)  return 0;  pair<int int> cur = make_pair(A B);  // if already calculated return calculated value  if (memo.find(cur) != memo.end())  return memo[cur];  int temp;  // step to areas on basis of last choose area  switch(last)  {  case 1:  temp = 1 + max(maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 2:  temp = 1 + max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 3:  temp = 1 + max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo));  break;  }  // store the result into map  memo[cur] = temp;  return temp; } // method returns maximum survival time int getMaxSurvivalTime(int A int B area X area Y area Z) {  if (A <= 0 || B <= 0)  return 0;  map< pair<int int> int > memo;  // At first we can step into any of the area  return  max(maxSurvival(A + X.a B + X.b X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b X Y Z 3 memo)); } // Driver code to test above method int main() {  area X(3 2);  area Y(-5 -10);  area Z(-20 5);  int A = 20;  int B = 8;  cout << getMaxSurvivalTime(A B X Y Z);  return 0; }

Java

/*package whatever //do not write package name here */ import java.util.*; import java.io.*; class GFG {  // Java code to get maximum survival time  // class to represent an area  static class area  {  // increment or decrement in A and B  public int a b;  public area(int a int b){  this.a = a;  this.b = b;  }  };  // class to represent pair  static class Pair{  public int firstsecond;  public Pair(int firstint second){  this.first = first;  this.second = second;  }  }  // Utility method to get maximum of 3 integers  static int max(int a int b int c)  {  return Math.max(a Math.max(b c));  }  // Utility method to get maximum survival time  static int maxSurvival(int A int B area X area Y area Z  int last HashMap<Pair Integer> memo)  {  // if any of A or B is less than 0 return 0  if (A <= 0 || B <= 0)  return 0;  Pair cur = new Pair(A B);  // if already calculated return calculated value  if (memo.containsKey(cur))  return memo.get(cur);  int temp = 0;  // step to areas on basis of last choose area  switch(last)  {  case 1:  temp = 1 + Math.max(maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 2:  temp = 1 + Math.max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 3:  temp = 1 + Math.max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo));  break;  }  // store the result into map  memo.put(curtemp);  return temp;  }  // method returns maximum survival time  static int getMaxSurvivalTime(int A int B area X area Y area Z)  {  if (A <= 0 || B <= 0)  return 0;  HashMap<PairInteger> memo = new HashMap<>();  // At first we can step into any of the area  return  max(maxSurvival(A + X.a B + X.b X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b X Y Z 3 memo));  }  // Driver Code  public static void main(String args[])  {  area X = new area(3 2);  area Y = new area(-5 -10);  area Z = new area(-20 5);  int A = 20;  int B = 8;  System.out.println(getMaxSurvivalTime(A B X Y Z));  } } // This code is contributed by shinjanpatra

Python3

# Python code to get maximum survival time # Class to represent an area class area: def __init__(self a b): self.a = a self.b = b # Utility method to get maximum survival time def maxSurvival(A B X Y Z last memo): # if any of A or B is less than 0 return 0 if (A <= 0 or B <= 0): return 0 cur = area(A B) # if already calculated return calculated value for ele in memo.keys(): if (cur.a == ele.a and cur.b == ele.b): return memo[ele] # step to areas on basis of last chosen area if (last == 1): temp = 1 + max(maxSurvival(A + Y.a B + Y.b X Y Z 2 memo) maxSurvival(A + Z.a B + Z.b X Y Z 3 memo)) elif (last == 2): temp = 1 + max(maxSurvival(A + X.a B + X.b X Y Z 1 memo) maxSurvival(A + Z.a B + Z.b X Y Z 3 memo)) elif (last == 3): temp = 1 + max(maxSurvival(A + X.a B + X.b X Y Z 1 memo) maxSurvival(A + Y.a B + Y.b X Y Z 2 memo)) # store the result into map memo[cur] = temp return temp # method returns maximum survival time def getMaxSurvivalTime(A B X Y Z): if (A <= 0 or B <= 0): return 0 memo = dict() # At first we can step into any of the area return max(maxSurvival(A + X.a B + X.b X Y Z 1 memo) maxSurvival(A + Y.a B + Y.b X Y Z 2 memo) maxSurvival(A + Z.a B + Z.b X Y Z 3 memo)) # Driver code to test above method X = area(3 2) Y = area(-5 -10) Z = area(-20 5) A = 20 B = 8 print(getMaxSurvivalTime(A B X Y Z)) # This code is contributed by Soumen Ghosh.

// C# code to get maximum survival time using System; using System.Collections.Generic; class GFG {  // class to represent an area  class area {  // increment or decrement in A and B  public int a b;  public area(int a int b)  {  this.a = a;  this.b = b;  }  };  // class to represent pair  class Pair {  public int first second;  public Pair(int first int second)  {  this.first = first;  this.second = second;  }  }  // Utility method to get maximum of 3 integers  static int max(int a int b int c)  {  return Math.Max(a Math.Max(b c));  }  // Utility method to get maximum survival time  static int maxSurvival(int A int B area X area Y  area Z int last  Dictionary<Pair int> memo)  {  // if any of A or B is less than 0 return 0  if (A <= 0 || B <= 0)  return 0;  Pair cur = new Pair(A B);  // if already calculated return calculated value  if (memo.ContainsKey(cur))  return memo[cur];  int temp = 0;  // step to areas on basis of last choose area  switch (last) {  case 1:  temp  = 1  + Math.Max(maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 2:  temp  = 1  + Math.Max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo));  break;  case 3:  temp  = 1  + Math.Max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo));  break;  }  // store the result into map  memo[cur] = temp;  return temp;  }  // method returns maximum survival time  static int getMaxSurvivalTime(int A int B area X  area Y area Z)  {  if (A <= 0 || B <= 0)  return 0;  Dictionary<Pair int> memo  = new Dictionary<Pair int>();  // At first we can step into any of the area  return max(  maxSurvival(A + X.a B + X.b X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b X Y Z 3  memo));  }  // Driver Code  public static void Main(String[] args)  {  area X = new area(3 2);  area Y = new area(-5 -10);  area Z = new area(-20 5);  int A = 20;  int B = 8;  Console.WriteLine(  getMaxSurvivalTime(A B X Y Z));  } } // This code is contributed by lokeshpotta20.

JavaScript

<script> // JavaScript code to get maximum survival time // Class to represent an area class area{  constructor(a b){  this.a = a  this.b = b  } } // Utility method to get maximum survival time function maxSurvival(A B X Y Z last memo){  // if any of A or B is less than 0 return 0  if (A <= 0 || B <= 0)  return 0  let cur = new area(A B)  // if already calculated return calculated value  for(let [keyvalue] of memo){  if (cur.a == key.a && cur.b == key.b)  return memo.get(key)  }  let temp;  // step to areas on basis of last chosen area  if (last == 1){  temp = 1 + Math.max(maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo))  }  else if (last == 2){  temp = 1 + Math.max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Z.a B + Z.b  X Y Z 3 memo))  }  else if (last == 3){  temp = 1 + Math.max(maxSurvival(A + X.a B + X.b  X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b  X Y Z 2 memo))  }  // store the result into map  memo.set(cur  temp)  return temp } // method returns maximum survival time function getMaxSurvivalTime(A B X Y Z){  if (A <= 0 || B <= 0)  return 0  let memo = new Map()  // At first we can step into any of the area  return Math.max(maxSurvival(A + X.a B + X.b X Y Z 1 memo)  maxSurvival(A + Y.a B + Y.b X Y Z 2 memo)  maxSurvival(A + Z.a B + Z.b X Y Z 3 memo)) } // Driver code to test above method let X = new area(3 2) let Y = new area(-5 -10) let Z = new area(-20 5) let A = 20 let B = 8 document.write(getMaxSurvivalTime(A B X Y Z)'
')   // This code is contributed by shinjanpatra </script>

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