UVa 590 - Always on the run

Summary
A thief wants to find the cheapest way of travelling to a certain city in exactly k days, given that each day she must make exactly one flight.

Explanation
Letting:


 * $$minCost[a][b]$$
 * be the minimal cost to travel to city $$a$$ on $$b$$ day
 * $$flightCost(i, j, d)$$
 * be the cost of the single flight from city $$i$$ to city $$j$$ on day $$d$$, where $$flightCost(i, j, d) = 0$$ denotes a non-existent flight.
 * $$n$$
 * be the number of possible cities the thief can escape to (including the currently inhibited one)
 * }
 * be the number of possible cities the thief can escape to (including the currently inhibited one)
 * }

We can deduce that:
 * $$ minCost[a][b] = min(minCost[i][b - 1] + flightCost(i, a, b)), \forall i: 1 \le i \le n, i \ne a, flightCost(i, a, b) \ne 0$$

Which is very similar to the classical UVa 116 problem.

The minimal cost can be calculated quickly using a bottom-up dynamic programming solution.

Input
3 6 2 130 150 3 75 0 80 7 120 110 0 100 110 120 0 4 60 70 60 50 3 0 135 140 2 70 80 2 3 2 0 70 1 80 0 0

Output
Scenario #1 The best flight costs 460.

Scenario #2 No flight possible.