kawatea's solution http://mirror.codeforces.com/contest/246/submission/2621642 i think it's brute force because this solution uesd a "for" loop to find distinct names of k sons
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kawatea's solution http://mirror.codeforces.com/contest/246/submission/2621642 i think it's brute force because this solution uesd a "for" loop to find distinct names of k sons
Название |
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It caches every query. It seems that there could not be many distinct large queries.
for (i = p; i < q; i++) { if (f[w[z][i].second] == 0) { sum++; f[w[z][i].second] = 1; } }
In the worst case this part works (summary) in
.
Sample:
K+M -> K+M-1 -> K+M-2 -> .... K
K -> 1,2,3, ..., K-1
K+1 -> some chain of length 1
K+2 -> some chain of length 2
...
K+M -> some chain of length M
Prove:
Each
times. To view it k times, we should ask about layer
.
w[z][i]
will be viewed at mostz
from k different ancestors ofw[z][i]
. How to get k different pairs <p,q>? For each of these k ancestors we have to add a chain from it to layerz
. Lengths of these chains will be at least 1,2, ... k. Sum is at most n. So k is at most