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/* |
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* Copyright (C) 2000-2004 Object Oriented Parallel Simulation Engine (OOPSE) project |
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* |
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* Contact: oopse@oopse.org |
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* |
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* This program is free software; you can redistribute it and/or |
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* modify it under the terms of the GNU Lesser General Public License |
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* as published by the Free Software Foundation; either version 2.1 |
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* of the License, or (at your option) any later version. |
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* All we ask is that proper credit is given for our work, which includes |
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* - but is not limited to - adding the above copyright notice to the beginning |
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* of your source code files, and to any copyright notice that you may distribute |
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* with programs based on this work. |
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* |
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* This program is distributed in the hope that it will be useful, |
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* but WITHOUT ANY WARRANTY; without even the implied warranty of |
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the |
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* GNU Lesser General Public License for more details. |
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* |
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* You should have received a copy of the GNU Lesser General Public License |
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* along with this program; if not, write to the Free Software |
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* Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. |
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/* |
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* Copyright (c) 2005 The University of Notre Dame. All Rights Reserved. |
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* |
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* The University of Notre Dame grants you ("Licensee") a |
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* non-exclusive, royalty free, license to use, modify and |
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* redistribute this software in source and binary code form, provided |
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* that the following conditions are met: |
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* |
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* 1. Acknowledgement of the program authors must be made in any |
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* publication of scientific results based in part on use of the |
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* program. An acceptable form of acknowledgement is citation of |
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* the article in which the program was described (Matthew |
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* A. Meineke, Charles F. Vardeman II, Teng Lin, Christopher |
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* J. Fennell and J. Daniel Gezelter, "OOPSE: An Object-Oriented |
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* Parallel Simulation Engine for Molecular Dynamics," |
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* J. Comput. Chem. 26, pp. 252-271 (2005)) |
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* |
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* 2. Redistributions of source code must retain the above copyright |
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* notice, this list of conditions and the following disclaimer. |
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* |
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* 3. Redistributions in binary form must reproduce the above copyright |
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* notice, this list of conditions and the following disclaimer in the |
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* documentation and/or other materials provided with the |
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* distribution. |
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* |
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* This software is provided "AS IS," without a warranty of any |
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* kind. All express or implied conditions, representations and |
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* warranties, including any implied warranty of merchantability, |
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* fitness for a particular purpose or non-infringement, are hereby |
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* excluded. The University of Notre Dame and its licensors shall not |
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* be liable for any damages suffered by licensee as a result of |
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* using, modifying or distributing the software or its |
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* derivatives. In no event will the University of Notre Dame or its |
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* licensors be liable for any lost revenue, profit or data, or for |
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* direct, indirect, special, consequential, incidental or punitive |
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* damages, however caused and regardless of the theory of liability, |
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* arising out of the use of or inability to use software, even if the |
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* University of Notre Dame has been advised of the possibility of |
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* such damages. |
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*/ |
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/** |
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* @file GenerateCombination.hpp |
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* @file next_combination.hpp |
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* @author tlin |
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* @date 10/27/2004 |
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* @version 1.0 |
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#include <vector> |
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#include <iterator> |
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#include <iostream> |
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namespace oopse { |
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/** |
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* @fn bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) |
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* @brief STL next_permuationtation like combination sequence generator. |
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* Given the first and last iterator of a sequence, next_combination iteratively generates all possible combination. |
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* Given the first and last iterator of a sequence, next_combination iteratively generates all |
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* possible combinations. |
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* @return if more combination is availiable, otherwise return false |
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* @param iterContainer iterator container |
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* @param first the first iterator |
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* @param last the last iterator |
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* @note first and last must be random access iterators and iterContainer must be the container which |
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* element is iterator. And all of the iteratos in iterContainer must be within the range [first, last) |
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* @note first and last must be random access iterators and iterContainer must be the container of |
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* random access iterators . And all of the iteratos in iterContainer must be within the range [first, last) |
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* |
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* @code |
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* std::vector<int> iv; |
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* //1 |
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* //8 |
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* //1 8 |
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* @endcode |
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*/ |
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template<class RandomAccessIterator, template<typename ELEM, typename = std::allocator<ELEM> > class IteratorContainer> |
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bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) { |
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template<class RandomAccessIterator> |
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bool next_combination(std::vector<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) { |
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if (first == last) { |
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return false; |
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} |
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RandomAccessIterator endIter = --last; |
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typename IteratorContainer<RandomAccessIterator>::iterator i = iterContainer.end(); |
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typename std::vector<RandomAccessIterator>::iterator i = iterContainer.end(); |
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if (iterContainer.empty()) { |
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//if sequence is empty, we insert the first iterator |
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iterContainer.insert(iterContainer.end(), first); |
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return true; |
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} else if (*(--i) != endIter){ |
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//if the last iterator in iterContainer does not reaches the end, just increment it |
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//if the last iterator in iterContainer does not reaches the end, just increase its iterator by 1 |
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++(*i); |
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return true; |
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} else {// the last iterator in iterContainer does not reaches the end |
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//starts at the end of the sequence and works its way towards the front, looking for two |
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//consecutive members of the sequence where the difference between them is greater |
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//than one. For example , if the sequence contains 1, 5, 8, 9 (total number is 10, begin |
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//index is 0, therefore 9 is the end index, and the current size is 4). At the end of while |
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//than one. For example , if the sequence contains 1, 5, 8, 9 (total number is 10, first is 0 |
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//and the last is 10 (due to STL's half open range)). At the end of while |
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//loop, j will point to 5, and i will point to 8, next combination should be 1, 6, 7, 8. |
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//If j is less than zero, it means it already reaches the last combination of current size. |
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//For instance, sequence may contain 6, 7, 8, 9 at this time, we need to increase the size |
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// of combination to 5 |
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typename IteratorContainer<RandomAccessIterator>::iterator j = i; |
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typename std::vector<RandomAccessIterator>::iterator j = i; |
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j--; |
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while( j >= iterContainer.begin() && *i == *j + 1){ |
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i--; |
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} |
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} //end next_combination |
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bool replaceWildCard(std::vector<std::vector<std::string>::iterator>& cont, |
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std::vector<std::string> sequence, std::vector<std::string> result |
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std::string& wildCard = "X") { |
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if (cont.size() > sequnce.size()) { |
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std::cerr << "the size of iterator container is greater than the size of sequence" |
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} |
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|
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bool hasMoreCombination = next_combination(cont, sequence.begin(), sequence.end()); |
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if (hasMoreCombination) { |
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result.resize(sequence.size()); |
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result.insert(result.end(), sequence.size(), wildCard); |
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for ( i = cont.begin(); i != cont.end(); i++){ |
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result[*i - sequence.begin()] = **i; |
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} |
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return hasMoreCombination; |
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} |
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} //end namespace oopse |
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#endif //UTILS_NEXT_COMBINATION_HPP |
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