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Example 3: Travelling Salesman problem
Given: n associated cities and distances among them
Find: tour of minimum length that visits all of city.
Solutions: How several tours are possible?
n*(n -1)...*1 = n!
Because n! > 2(n-1)
Therefore n! = ? (2n) (lower bound)
As of now, there is no algorithm that determines a tour of minimum length plus covers all of the cities in polynomial time. But, there are many very good heuristic algorithms.
Implement multiple queues in a single dimensional array. Write algorithms for various queue operations for them.
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code for count and display
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