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.

// 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 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 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; }

Output:

5

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