USACO Training: Job Processing

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It's about time I finished section 4.2 of USACO training...

USACO Training: Job Processing


We can compute the first variable of output by attempting to greedily assign the tasks to the type-A machines. This is completed by keeping track of both
  1. how much time an machine spends on finishing an arbitrary task AND
  2. how much time the machine ALREADY SPENT on finishing assigned TASKS.
The first comes from input; the second is initially zero and gradually increases.
We wish to minimize the time it takes to finish task N. The time it will take to finish any task i is the minimal possible value of m.1 + m.2 where m is any machine. Once the machine m is selected such that this value is minimized, increase the m.2 value of this machine by m.1. We repeat this process for all i from 1 to N. Notice that the answer is not dependent on what i is, but only on the m.1 and m.2 values.

Assume we store these minimum values in an array, T[1] to T[N]. Clearly the first answer is T[N]. Now, we observe that T[1] to T[N] is always nondecreasing, i.e. for arbitrary T[i], T[i] <= T[i+1]. 

Proof: Suppose otherwise, that T[i+1] < T[i]. Hence T[i+1] "found" a minimum m.1 + m.2 that was smaller than what T[i] "found". Since this value exists, then T[i] would have found this value instead, leading to a contradiction.

Now, for the B machines, let's start from N and go down to 1 instead, finding the minimum m.1 + m.2 for the B machines. Store the minimums in an array U[1] to U[N]. If the B machines were independent from the A machines, then U[1] would be the amount of time the B machines took. Notice that for the same reason T[1] to T[N] is always nondecreasing, U[N] to U[1] is always nondecreasing, or equivalently, U[1] to U[N] is always nonincreasing. 

The maximum value of T[i] + U[i] will be the answer, since this virtually means "all the time B spends on task i plus all the time B spends waiting for task i."

By the greedy load balancing problem (my last post), the value max(T[i]+U[i]) is minimized this way, hence, this value is optimal.

And now we have the two answers! 

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