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вернуться в форумIt was very easy. Backtracking(перебор с возвратом) I think that it is hard problem and has exp(n) complexity. If remove psevdopractic decoration it is to solve a system boolean equation of 100 unknows Xi with type of Xi^Xj=0; (NotXi)^Xj=0;(NotXi)^(NotXj)=0. Amount of equation is near 10000. Thus we have easy problem for weak tests and very hard problem for detailed test. This situation was brightly shoun for identical Ships problems.Programmers should create code working on all possible tests in prescribed range of variables. Now I am also having Ac(0.031) by using backtracking. I have applied this method to boolean problem not to Graf. But I fear that we all will lost our submits if problem will be rejudged. In worst case in complexy is O(n^2) Yes, it's O(N^2). And resembles another problem of this type: 1382 does transitive closure algorithm work here? Just a standard 2-SAT problem. |
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