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, 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.
CPP
// C++ code to get maximum survival time#include <bits/stdc++.h>using namespace std; // structure to represent an areastruct 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 integersint max(int a, int b, int c){ return max(a, max(b, c));} // Utility method to get maximum survival timeint 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 chose 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 timeint 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 methodint 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;} |
Python3
# Python code to get maximum survival time # Class to represent an areaclass area: def __init__(self, a, b): self.a = a self.b = b # Utility method to get maximum survival timedef 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 timedef 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 methodX = area(3, 2)Y = area(-5, -10)Z = area(-20, 5) A = 20B = 8print(getMaxSurvivalTime(A, B, X, Y, Z)) # This code is contributed by Soumen Ghosh. |
Output:
5
This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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