Catch the Plane

This problem asks gives us a bus schedule, where each bus has a probability of ACTUALLY running, and then asks us to figure out the max probability we can arrive at a given node before a given time.

Standard problems might ask us to optimize for the probability of ever reaching a given location, or the minimum time it takes to get there, but attempting to do both at the same time makes this slightly more difficult. A way to fix this might be to use a dp:

dp[node][time] = maximum probability of getting to node at time

or swapping the dimensions

dp[node][probability] = minimum time of getting to a node with at least a given probability

The issue with the first one is time is 10^18. Even with , where wee only consider times that a bus leaves or arrives at, we are still at 10^6^2 = 10^12. The second method is not viable since probability doesn't give a discrete measure, or perhaps more validly, there are too many discrete probabilities to memoize. Plus, it's not inherently clear what the relation would look like.

So lets go with some fundamentals and see where it gets us:

The probability of reaching the the airport by taking a bus is the probability we get there by taking our ideal first bus plus the probability that bus doesn't run times the probability of getting there on some later bus.

Based on this, we see that the probability of getting there depends on

1. Our choice of bus to take
2. ultimately the probability of getting there on the LAST possible bus

So when we arrive at a node, we would have to recurse on the LAST bus to leave the stop, then recurse on the second to last one, calculate whether it makes sense to try to take that second to last bus, and then get a probability of getting there on EITHER of the last two buses (max(p(last), p(second_to_last) * p(second_to_last_runs) + p(last) * 1-p(second_to_last_runs))). We can then repeat the process for each earlier bus that leaves the stop. This will get us an answer, but note that we have done O(n) work for this bus. What happens when we get a slightly later bus that arrives at this station? Do we have to calculate this work all over again? No. The answers to p(last) and p(second_to_last) should be the same. Even performing the calculation again with cached values for each p, however, is enough to add that factor of O(n). To avoid this, at each node, we have to cache a list for each departing bus, namely, what is the probability of reaching if I arrive before THIS bus? This is calculated the first time we arrive at a node, and then simply looked up (be sure to sort it for fast lookup! or just use a treeset based on time) at each arrival. When a bus arrives, look up the probability we get there based on the next time a bus leaves (note that this may not involve actually TAKING this bus, but that doesn't matter). The code looks as follows:

double goes[]; // probability a bus actually goes
double memo[]; // probability of arriving on time if we go on this bus
treeset<Pair>[] memo2; // array, one for each node, of a treeset of pairs. pair.first = time. pair.second = probability. sort based on time, and probability represents best probability after that time
recurse(bus) {
    if (memo[bus] != inf) return memo[bus];
    temp = memo2[bus.arrival_station];
    for (i : buses leaving bus.arrival_station in reverse order) { //
        recurse(i);
        p = max(temp.get((i+1).leave_time), memo[i]*goes[i] + temp.get((i+1).leave_time)*(1-goes[i])) // don't try to take i, OR try to take i, and it either goes or it doesn't
        temp.add(i.leave_time, p)
    }
    memo[bus] = temp.ciel(bus.arrive_time); //returns the first element with a time greater than our input
    return memo[bus];
}

While this solution is correct and fast, the issue is the recursion depth. With 10^6 buses and possible stack frames, we're going to smash that. So naturally, we simply convert this to a DP problem