Hi,
I am trying to solve this problem in spoj website. you might be
aware of this website.
http://www.spoj.com/problems/GERGOVIA
Gergovia consists of one street, and every inhabitant of the city is a wine
salesman. Everyone buys wine from other inhabitants of the city. Every day
each inhabitant decides how much wine he wants to buy or sell.
Interestingly, demand and supply is always the same, so that each
inhabitant gets what he wants.
There is one problem, however: Transporting wine from one house to another
results in work. Since all wines are equally good, the inhabitants of
Gergovia don't care which persons they are doing trade with, they are only
interested in selling or buying a specific amount of wine.
In this problem you are asked to reconstruct the trading during one day in
Gergovia. For simplicity we will assume that the houses are built along a
straight line with equal distance between adjacent houses. Transporting one
bottle of wine from one house to an adjacent house results in one unit of
work.
Input
The input consists of several test cases.
Each test case starts with the number of inhabitants N (2 ≤ N ≤ 100000).
The following line contains n integers ai (-1000 ≤ ai ≤ 1000).
If ai ≥ 0, it means that the inhabitant living in the ith house wants to
buy ai bottles of wine. If ai < 0, he wants to sell -aibottles of wine.
You may assume that the numbers ai sum up to 0.
The last test case is followed by a line containing 0.
Output
For each test case print the minimum amount of work units needed so that
every inhabitant has his demand fulfilled.
I tried but failed. I came across solutions which say we need to greedily
choose houses which satisfy our requirement from either left end or right
end. But I do not find good reasoning or proof for this approach. There are
few comments asking for explanation. But they do not have any response.
Suppose houses have wine requirement like this -1,-1,2,-1,1. In this case
if House with 2 wine bottles requirement chooses greedily, we get cost to
be 2 plus cost to transfer wine bottle from left edge to right edge. (2+4
=6). But we can see if we choosing greedily from one end gives cost 2+1+1
= 4. Why greedy approach from one end succeeds while choosing arbitrarily
fails.
Any hints or proofs is greatly appreciated.
--
-Thanks
Bujji
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