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, thatās true, good point! Iām not sure why I defaulted to random, but I guess itās 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 donāt really need it.
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because using long long increases time consumption so i think your solution got tle
Can anybody explain how does the radius of search become half ?
Can we transform the initial graph into minimum spanning tree and using dynamic programming to figure out the minimum distance between two special nodes?Iāve tried that but get WA and I cannot think of any test case that can prove my idea is wrong.Can you guys please help me?
my approach is same as the editorialā¦ what is wrong with the solution: https://paste.ubuntu.com/p/bfQccvhGhd/
