UVa 590 - Always on the run

590 - Always on the run

 * http://acm.uva.es/p/v5/590.html

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
Let cost[a][b] be the cost to travel to city a on b day.

cost[a][b] could be calculated as the minimum of cost[m][b-1] + cost to travel from city m to city a.

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.