====================================================================================================
Q: Can vectors be used instead of arrays.

A: OK. Use it. But you need vector of vectors. Fine, go ahead. I believe arrays are much simpler.
====================================================================================================
Q: Can we use global variables?

A: Use of global variables is very deprecated. Avoid those. Not only in this assignment, but
   throughout the rest of your life.
====================================================================================================
Q: Can we use C++ classes?

A: C++ should be used for classes only. You may have classes (like digraph). But do not change the
   data organization in your classes from what is given in the assignment statement.
====================================================================================================
Q: How to print graphs and charts?

A: Use the format specified in the sample I/O section. You may write functions for printing graphs
   and charts. In fact, writing functions is a good idea because printing of graphs/charts is done
   multiple times.
====================================================================================================
Q: More clarification for Chart 2 (Part 4).

A: Use the O(n^4) algorithm given in the statement, not the Floyd--Warshall-based O(n^5) algorithm.
====================================================================================================
Q: More clarification for Chart 3 (Part 5).

A: If C1(i,j) < infinity (that is, there is an AI path in the red graph from i to j), take
   C3(i,j) = C1(i,j). If C1(i,j) = infinity, choose the shortest path with blue edges on it. No
   restrictions on how many blue edges will be there in the shortest paths. In this part, you are
   not asked to maximize the number of AI legs (or minimize the number of non-AI legs). You are
   asked to minimize the cost by including red and blue edges in the best manner.
====================================================================================================
Q: PART4 it is given choose atmost one non ai leg but in the sample output given 1->4 and 4->0 two
   are chosen why?

A: In the sample output, 1 -> 0 cost is shown as -. Individually, 1 -> 4 and 4 -> 0 travels are now
   possible. But in the last part, 1 -> 0 cost is shown as 9751 (because two blue edges are
   allowed in Part 5).
====================================================================================================
Q: in part4 single (k,l) non-ai flight does not cover all (i,j)s, so how do we decide which
   uncovered (i,j) should be considered?

A: For each uncovered (i,j), minimize over ALL blue edges (k,l). Some blue edge (k,l) may have no
   impact on the given (i,j). This does not matter. Check all with the hope that some of these
   might work.

   More clarification:

   For each (i,j), if C1(i,j) = INFINITY, then:
      For all blue edges (k,l), compute possibility[i,j](k,l) = C1(i,k) + edgecost(k->l) + C1(l,j).
      Take C2(i,j) = min(possibility[i,j](k,l)), where the minimum is over all blue edges (k,l).

   Notes

      (1) For different AI-uncovered (i,j) pairs, the minimum may be achieved by different blue
          edges (k,l).

      (2) For some AI-uncovered (i,j) pairs, you may still have C2(i,j) = INFINITY (no blue edge
          (k,l) helps).
====================================================================================================
Q: What is the maximum value of n I should prepare for?

A: 32.
====================================================================================================
Q: What can be the maximum cost?

A: Each single leg costs < 10000. If n <= 32, the maximum price can be 310,000.
====================================================================================================