Mercurial > hg > orthanc-stone
view OrthancStone/Resources/Graveyard/RTStructTentativeReimplementation-BGO/DisjointDataSet.h @ 2034:4b24b7533346 deep-learning
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author | Sebastien Jodogne <s.jodogne@gmail.com> |
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date | Fri, 27 Jan 2023 15:43:24 +0100 |
parents | affde38b84de |
children | 07964689cb0b |
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/** * Stone of Orthanc * Copyright (C) 2012-2016 Sebastien Jodogne, Medical Physics * Department, University Hospital of Liege, Belgium * Copyright (C) 2017-2022 Osimis S.A., Belgium * Copyright (C) 2021-2022 Sebastien Jodogne, ICTEAM UCLouvain, Belgium * * This program is free software: you can redistribute it and/or * modify it under the terms of the GNU Lesser General Public License * as published by the Free Software Foundation, either version 3 of * the License, or (at your option) any later version. * * This program is distributed in the hope that it will be useful, but * WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU * Lesser General Public License for more details. * * You should have received a copy of the GNU Lesser General Public * License along with this program. If not, see * <http://www.gnu.org/licenses/>. **/ #pragma once #include <vector> #include "../StoneException.h" namespace OrthancStone { class DisjointDataSet { public: DisjointDataSet(size_t itemCount) : parents_(itemCount), ranks_(itemCount) { for (size_t index = 0; index < parents_.size(); index++) { SetParent(index,index); ranks_[index] = 1; } } size_t Find(size_t item) { /* If parents_[i] == i, it means i is representative of a set. Otherwise, we go up the tree... */ if (GetParent(item) != item) { // if item is not a top item (representative of its set), // we use path compression to improve future lookups // see: https://en.wikipedia.org/wiki/Disjoint-set_data_structure#Path_compression SetParent(item, Find(parents_[item])); } // now that paths have been compressed, we are positively certain // that item's parent is a set ("X is a set" means that X is the // representative of a set) return GetParent(item); } /* This merge the two sets that contains itemA and itemB */ void Union(size_t itemA, size_t itemB) { // Find current sets of x and y size_t setA = Find(itemA); size_t setB = Find(itemB); // if setA == setB, it means they are already in the same set and // do not need to be merged! if (setA != setB) { // we need to merge the sets, which means that the trees representing // the sets needs to be merged (there must be a single top parent to // all the items originally belonging to setA and setB must be the same) // since the algorithm speed is inversely proportional to the tree // height (the rank), we need to combine trees in a way that // minimizes this rank. See "Union by rank" at // https://en.wikipedia.org/wiki/Disjoint-set_data_structure#by_rank if (GetRank(setA) < GetRank(setB)) { SetParent(setA, setB); } else if (GetRank(setA) > GetRank(setB)) { SetParent(setB, setA); } else { SetParent(setB, setA); BumpRank(setA); // the trees had the same height but we attached the whole of setB // under setA (under its parent), so the resulting tree is now // 1 higher. setB is NOT representative of a set anymore. } } } private: size_t GetRank(size_t i) const { ORTHANC_ASSERT(i < ranks_.size()); ORTHANC_ASSERT(ranks_.size() == parents_.size()); return ranks_[i]; } size_t GetParent(size_t i) const { ORTHANC_ASSERT(i < parents_.size()); ORTHANC_ASSERT(ranks_.size() == parents_.size()); return parents_[i]; } void SetParent(size_t i, size_t parent) { ORTHANC_ASSERT(i < parents_.size()); ORTHANC_ASSERT(ranks_.size() == parents_.size()); parents_[i] = parent; } void BumpRank(size_t i) { ORTHANC_ASSERT(i < ranks_.size()); ORTHANC_ASSERT(ranks_.size() == parents_.size()); ranks_[i] = ranks_[i] + 1u; } /* This vector contains the direct parent of each item */ std::vector<size_t> parents_; /* This vector contains the tree height of each set. The values in the vector for non-representative items is UNDEFINED! */ std::vector<size_t> ranks_; }; }