问题
I have a problem simlar to Huffman's encoding, I'm not sure exactly how it can be solved or if it is a reverse Huffman's encoding. But it definitely can be solved using a greedy approach.
Consider a set of length, each associated with a probability. i.e.
X={a1=(100,1/4),a2=(500,1/4),a3=(200,1/2)}
Obviously, the sum of all the probabilities = 1.
Arrange the lengths together on a line one after the other from a starting point.
For example: {a2,a1,a3} in that order from start to finish.
Define the cost of an element a_i as its the total length from the starting line to the end of this element multiplied by its probability.
So from the previous arrangement:
cost(a2) = (500)*(1/4)
cost(a1) = (500+100)*(1/4)
cost(a3) = (500+100+200)*(1/2)
Define the total cost as the sum of all costs. e.g. cost(X) = cost(a2) + cost(a1) + cost(a3). Give an algorithm that finds an arrangement that minimizes cost(X)
I've tried forming some alternative huffman trees but it doesn't work.
Sorting by probability will fail (consider X={(100,0.4),(300,0.6)}).
Sorting by length will also fail (consider X={(100,0.1),(300,0.9)}).
If anyone can help or hint towards an optimal solution algorithm, it would be great.
回答1:
Consider what happens if you swap two adjacent elements. The calculations before and after the two elements are the same, so it just depends on the two elements.
Taking two elements in isolation, the costs are P1L1 + P2(L1 + L2) and P2L2 + P1(L1 + L2). If you subtract this and simplify if I have got the algebra right you want to swap 1 to first when L1/P1 < L2/P2. Check - this at least gets the right answer when L1 = 0.
So I think you want to sort the elements into increasing order of Li/Pi, because if that is not the case you can improve the answer by swapping adjacent elements.
来源：https://stackoverflow.com/questions/46615768/reverse-huffmans-algorithm