Wednesday, December 19, 2012


SPOJ: 348. Expedition

Nice problem! The first observation is, if you want to reach the city, say, point 0, you have to ensure that every single point between the current position and city must also be reachable. Now, the task is to minimize the number of stoppages for fuel, which is at most 10000. So, we sort the fuel stations, and start from current position. For every fuel station, if we want to reach it, we must have fuel f more than or equal to the distance d. Also, using the larger capacities will always reduce the number of stations we must stop.

So, for each stoppage, starting from farthest, if we can reach this stoppage with existing fuel, we push it in a priority queue (max heap in this case) for future use. If we cannot reach a particular stoppage, then we keep popping the queue and keep adding the amounts with currently available until we are able to reach current stoppage, and then pushing this value. This strategy ensures an optimal solution to reach a particular stoppage, as the priority queue will hold maximum capacities seen so far along the path, but not used yet.

If at any time, the queue gets empty and the amount of fuel is not sufficient, then the particular stoppage cannot be reached, hence, it will be impossible to reach the city.

Happy Coding

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