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59 lines
1.9 KiB
Python
59 lines
1.9 KiB
Python
### PROBLEM ###
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"""
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We are given a rod of length n and we are given the array of prices, also of
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length n. This array contains the price for selling a rod at a certain length.
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For example, prices[5] shows the price we can sell a rod of length 5.
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Generalising, prices[x] shows the price a rod of length x can be sold.
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We are tasked to find the optimal solution to sell the given rod.
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"""
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### SOLUTION ###
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"""
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Profit(n) = max(1<i<n){Price(n),Price(i)+Profit(n-i)}
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When we receive a rod, we have two options:
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a) Don't cut it and sell it as is (receiving prices[length])
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b) Cut it and sell it in two parts. The length we cut it and the rod we are
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left with, which we have to try and sell separately in an efficient way.
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Choose the maximum price we can get.
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"""
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def CutRod(n):
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if(n == 1):
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#Cannot cut rod any further
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return prices[1]
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noCut = prices[n] #The price you get when you don't cut the rod
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yesCut = [-1 for x in range(n)] #The prices for the different cutting options
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for i in range(1,n):
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if(solutions[i] == -1):
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#We haven't calulated solution for length i yet.
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#We know we sell the part of length i so we get prices[i].
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#We just need to know how to sell rod of length n-i
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yesCut[i] = prices[i] + CutRod(n-i)
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else:
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#We have calculated solution for length i.
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#We add the two prices.
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yesCut[i] = prices[i] + solutions[n-i]
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#We need to find the highest price in order to sell more efficiently.
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#We have to choose between noCut and the prices in yesCut.
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m = noCut #Initialize max to noCut
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for i in range(n):
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if(yesCut[i] > m):
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m = yesCut[i]
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solutions[n] = m
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return m
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### EXAMPLE ###
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length = 5
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#The first price, 0, is for when we have no rod.
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prices = [0, 1, 3, 7, 9, 11, 13, 17, 21, 21, 30]
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solutions = [-1 for x in range(length+1)]
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print(CutRod(length))
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