@pushkarmishra probably because the author’s solution seems to be using a different input/output than the final problem? This is probably due to a revision of the problem. I think the int aa overflowed to a negative hence aa<2. Try another accepted solution? (like AnonymousBunny)
Thank you 
The first part of the solution is really nice. I think you can leave out the random part by choosing the sets appropriately. log(k)+1 should be enough by e.g using the ith special point for the starting set in round j if the jth digit in the binary representation of i is one.
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@ceilks Yeah, that’s true, good point! I’m not sure why I defaulted to random, but I guess it’s kind of nice since you could see it as log(# of tests) with the full feedback.
That is because handling of 64 bit integers is slower than 32 bit integers. Thus, you should avoid long long whenever you don’t really need it.
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because using long long increases time consumption so i think your solution got tle
Nice one man
Nice one!!!
Alternative solution: optimised version of subtask 2
while (!pq.empty() && pq.top().ff < ans)
Here, we break when we’ve exceeded the current best distance
if (node.second != source && node.isSpecial()) ans = min(ans,node.first);
And here we minimise the distance the first time we hit a special node (and then break)