📜  区域选择

📅  最后修改于: 2021-09-22 10:09:54             🧑  作者: Mango

考虑一个游戏,你有两种类型的力量,A 和 B,有 3 种类型的区域 X、Y 和 Z。每一秒你必须在这些区域之间切换,每个区域都有特定的属性,你的力量 A 和功率 B 增加或减少。我们需要不断选择区域,以使我们的生存时间最大化。当任何幂,A 或 B 达到小于 0 时,生存时间结束。
例子:

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

这个问题可以使用递归来解决,在每个时间单元之后我们可以去任何一个区域,但我们会选择最终导致最大生存时间的区域。由于递归会导致多次求解相同的子问题,我们将根据 A 和 B 的幂记忆结果,如果我们达到同一对 A 和 B 的幂,我们将不再求解,而是采用先前计算的结果。
下面给出了上述方法的简单实现。

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, int>& memo)
{
    //  if any of A or B is less than 0, return 0
    if (A <= 0 || B <= 0)
        return 0;
    pair 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 > 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;
}


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.


输出:

5

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