📜  用于切割棒的Python程序 | DP-13

📅  最后修改于: 2022-05-13 01:56:56.533000             🧑  作者: Mango

用于切割棒的Python程序 | DP-13

给定一根长度为 n 英寸的杆和一个价格数组,其中包含所有尺寸小于 n 的块的价格。确定通过切割杆并出售碎片可获得的最大值。例如,如果杆的长度为 8,并且不同部分的值如下所示,则最大可获得值为 22(通过切割长度为 2 和 6 的两根)

length   | 1   2   3   4   5   6   7   8  
--------------------------------------------
price    | 1   5   8   9  10  17  17  20

如果价格如下,那么最大可获得值为24(通过切割8块长度1)

length   | 1   2   3   4   5   6   7   8  
--------------------------------------------
price    | 3   5   8   9  10  17  17  20

以下是棒切割问题的简单递归实现。实现简单地遵循上面提到的递归结构。

Python3
# A Naive recursive solution
# for Rod cutting problem
import sys
 
# A utility function to get the
# maximum of two integers
def max(a, b):
    return a if (a > b) else b
     
# Returns the best obtainable price for a rod of length n
# and price[] as prices of different pieces
def cutRod(price, n):
    if(n <= 0):
        return 0
    max_val = -sys.maxsize-1
     
    # Recursively cut the rod in different pieces 
    # and compare different configurations
    for i in range(0, n):
        max_val = max(max_val, price[i] +
                      cutRod(price, n - i - 1))
    return max_val
 
# Driver code
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is", cutRod(arr, size))
 
# This code is contributed by 'Smitha Dinesh Semwal'


Python3
# A Dynamic Programming solution for Rod cutting problem
INT_MIN = -32767
 
# Returns the best obtainable price for a rod of length n and
# price[] as prices of different pieces
def cutRod(price, n):
    val = [0 for x in range(n + 1)]
    val[0] = 0
 
    # Build the table val[] in bottom up manner and return
    # the last entry from the table
    for i in range(1, n + 1):
        max_val = INT_MIN
        for j in range(i):
             max_val = max(max_val, price[j] + val[i-j-1])
        val[i] = max_val
 
    return val[n]
 
# Driver program to test above functions
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is " + str(cutRod(arr, size)))
 
# This code is contributed by Bhavya Jain


输出:
Maximum Obtainable Value is 22

考虑到上述实现,以下是长度为 4 的 Rod 的递归树。

cR() ---> cutRod() 

                             cR(4)
                  /        /           
                 /        /              
             cR(3)       cR(2)     cR(1)   cR(0)
            /  |         /         |
           /   |        /          |  
      cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
     /        |          |
    /         |          |   
  cR(1) cR(0) cR(0)      cR(0)
   /
 /
CR(0)

在上面的部分递归树中,cR(2) 被求解了两次。我们可以看到有很多子问题被一次又一次地解决。由于再次调用相同的子问题,此问题具有重叠子问题属性。因此,Rod Cutting 问题具有动态规划问题的两个属性(参见 this 和 this)。与其他典型的动态规划(DP)问题一样,通过以自下而上的方式构造临时数组 val[] 可以避免重复计算相同的子问题。

Python3

# A Dynamic Programming solution for Rod cutting problem
INT_MIN = -32767
 
# Returns the best obtainable price for a rod of length n and
# price[] as prices of different pieces
def cutRod(price, n):
    val = [0 for x in range(n + 1)]
    val[0] = 0
 
    # Build the table val[] in bottom up manner and return
    # the last entry from the table
    for i in range(1, n + 1):
        max_val = INT_MIN
        for j in range(i):
             max_val = max(max_val, price[j] + val[i-j-1])
        val[i] = max_val
 
    return val[n]
 
# Driver program to test above functions
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is " + str(cutRod(arr, size)))
 
# This code is contributed by Bhavya Jain
输出:
Maximum Obtainable Value is 22

请参阅有关切割棒的完整文章 | DP-13 了解更多详情!