Q as a Fibonacci heap In this case, the amortized cost of each of | V | DELETE-MIN operations is O(log
2
V).
Operation INSERT-KEY in the subroutine RELAX now takes only O(1)amortized time for each of the | E |
edges.
Let H denote the data structure which stores the partial solution as it is built up. The algorithm involves
| E | = m insert operations on H and | V |
deletemin operations involved in selecting the next node v from
V. So, the running time depends on the implementation of the heap H. Hence we need to discuss this
implementation.
In all heap implementations, we assume that we have an array of pointers ...
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