Consider a game, in which you have two types of powers, A and B and there are 3 types of Areas X, Y and Z. Every second you have to switch between these areas, and each area has specific properties by which your power A and power B increase or decrease. We need to keep choosing areas in such a way that our survival time is maximized. Survival time ends when any of the powers, A or B reaches less than 0.
Examples:
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
This problem can be solved using recursion, after each time unit we can go to any of the area but we will choose that area which ultimately leads to maximum survival time. As recursion can lead to solving same subproblem many time, we will memoize the result on basis of power A and B, if we reach to same pair of power A and B, we won’t solve it again instead we will take the previously calculated result.
Given below is the simple implementation of above approach.
// C++ code to get maximum survival time #include <bits/stdc++.h> 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;
} |
/*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 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, 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(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 ;
HashMap<Pair,Integer> 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 |
# 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. |
<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 [key,value] 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), "</br>" )
// This code is contributed by shinjanpatra </script> |
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