Dynamic programming problem for minimum cost [closed] - algorithm

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I have a cell tower question. There are n towns. We want to build cell tower in some of the towns. Each cell tower can cover itself and its neighbor. Each town has a cost to build cell tower. We want to find out the minimum cost to build the cell tower to cover all towns.
For example,
(1)
TOWN 1 2 3
COST 5 1 2
We select to build cell tower in town-2. The cost is 1.
(2)
TOWN 1 2 3 4
COST 5 1 2 3
We select to build cell tower in town-2/3. The cost is 1+2=3.
(3)
TOWN 1 2 3 4
COST 5 1 3 2
We select to build cell tower in town-2/4. The cost is 1+2=3.
It's a dynamic programming algorithm. How can I solve it?
Thanks
Ling

Answer 1

I'd go with something among the following lines:
f(0,_) = 0
f(1,true) = 0
f(1,false) = cost[1]
f(x,true) = min{ f(x-1,true) + cost[x], f(x-1,false) }
f(x,false) = min { f(x-1,true) + cost[x], f(x-2,true) + cost[x-1]}
The idea is:
x is the current number of city we are looking at, and the boolean is true if this city is already covered (by the city from the left).
f(0,_) is an empty base clause - it is free to cover nothing.
f(1,false) is base where city 1 is not covered, so you must put a tower there, and f(1,true) is a base where city 1 is covered, so you don't need a tower and you are done.
f(x,true) is the case that the city x is already covered, so you can put there a tower, and continue to the next city [which is also covered - f(x-1,true)], or don't put a tower there, and the next city is not covered [f(x-1,false)]
f(x,false) is the case where city x is not covered, so you basically have two choices, or put a tower in there, and then continue to f(x-1,true). Or put a tower in the next city (in x-1) and then continue to f(x-2,true)
Start with f(x,false), where x is the last city, and you'll get the minimal solution.
It is easy to see that this recursive formula can easily be modified to DP.