Resolved: It was using proxy settings from Firefox and not the system settings!

No, that doesn't work. There are cases where you may want to choose some other cell.

For each tree, only the last cut affects the final tree height. So we will assign to each tree the last cut (when, which device to cut that tree). We can solve it by max matching min cost (We have N trees and (number of devices)*time suggesions for last cut). Construct this graph is not difficult, but it will too slow as the number of edges is about 150^3. So the key idea to optimize is: For each tree, you know which last cut is good, which is bad (which makes the final tree height smaller is better). And you only need consider ~200 best last cut. The number of edges is now about 150^2.

I'll try to explain solution without flows.

If tree doesn't need to be cut it adds to answer value h[i] + T * a[i]

If it cuts in turn t from the end on device j, the value is dev[j] + t * a[i].

(Here you can see, if set of trees and devices used on some turn is fixed, order of matching doesn't affect the result).

Consider subset of trees that are going to be cut. Parameter h[i] significant no more. Now we can sort trees using parameter a[i] and say that we can cut trees only in that order.

On the other hand, if we consider one turn, and look on set of devices that are used on this turn, its obvious that optimal way to use subset with the smallest values of dev[j]. So we can sort all the devices and say that we are using some its prefix on each turn. So the last what is needed is to find how much devices should be used on each turn.

Now you can see the following dp can be used to solve this problem. dp[t][i][j] — minimal value for cutting i trees in t turns, where j is the number of devices used on current turn.

from the state (t, i, j) we can go to:

(t, i + 1, j) //do not cut tree i, costs h[i] + T * a[i]

(t, i + 1, j + 1) // cut tree i with device j with value dev[j] + t * a[i]

(t + 1, i, 0) // step in the next turn

The answer is the cheapest path from state (0, 0, 0) to (T, n, 0).

Maybe, you use cin.tie(0) and ios::sync_with_stdio(false)?

All of Alice's picks are equivalent, say she picks 0. For every number of points P the host says section 0 has, you need to find best section B for Bob to choose.

Expected value if you choose B is:

For every turn:
   thisprob = probability that 0 is picked on this turn

   expected += thisprob * expected increase of B for every subsegment that includes 0
   expected += (1 - thisprob) * expected increase of B for every subsegment that does not include 0

Then calculate this for all Bs and have Bob pick the best one. Calculating the actual probabilities is just a simple DP exercise, but if you can't figure out how to do it you can check my code on the practice room (I failed during the contest because of MLE :'( )

You had to do a binary search on the answer. Description of the check function: First construct a new graph, where a edge has 0 cost if road's height is more than king's shoe, else it is equal to (diff)^2 where diff=(height of king's shoe)-(road height) --> that is the minimum you would have to pay in case you want to traverse that road. Then as the number of nodes were small (50 at max.), you can apply floyd-warshall, to find the minimum cost to go from node '0' to node 'n-1'. If this cost <=king's budget, then the size of shoe (which is the argument of this bool function) is valid so return true, else false.

You only use the device with height 50. At time=2 cut the first tree, so in the end it will have height=50. At time=1, cut the second tree, so in the end it will have height=50+30.