Choice of Area

2.8

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

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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