考虑一个游戏，其中您拥有两种类型的力量，即A和B，并且存在三种类型的区域X，Y和Z。您必须每秒在这些区域之间进行切换，每个区域具有特定的属性，您的能力A和B功率B增大或减小。我们需要继续选择区域，以使我们的生存时间最大化。生存时间在A或B的任何一项力量小于0时结束。
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
可以使用递归来解决此问题，在每个时间单位之后，我们可以转到任何区域，但我们将选择最终导致最长生存时间的区域。由于递归可以导致多次解决相同的子问题，因此我们将基于功效A和B记住结果，如果我们达到相同的功效A和B对，我们将不再解决它，而是取先前计算的结果。
下面给出的是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