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root/group/branches/new_design/OOPSE-3.0/src/utils/next_combination.hpp
Revision: 1823
Committed: Thu Dec 2 02:23:45 2004 UTC (19 years, 7 months ago) by tim
File size: 5152 byte(s)
Log Message:
begin to fix linking problem

File Contents

# Content
1 /*
2 * Copyright (C) 2000-2004 Object Oriented Parallel Simulation Engine (OOPSE) project
3 *
4 * Contact: oopse@oopse.org
5 *
6 * This program is free software; you can redistribute it and/or
7 * modify it under the terms of the GNU Lesser General Public License
8 * as published by the Free Software Foundation; either version 2.1
9 * of the License, or (at your option) any later version.
10 * All we ask is that proper credit is given for our work, which includes
11 * - but is not limited to - adding the above copyright notice to the beginning
12 * of your source code files, and to any copyright notice that you may distribute
13 * with programs based on this work.
14 *
15 * This program is distributed in the hope that it will be useful,
16 * but WITHOUT ANY WARRANTY; without even the implied warranty of
17 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
18 * GNU Lesser General Public License for more details.
19 *
20 * You should have received a copy of the GNU Lesser General Public License
21 * along with this program; if not, write to the Free Software
22 * Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
23 *
24 */
25
26 /**
27 * @file next_combination.hpp
28 * @author tlin
29 * @date 10/27/2004
30 * @version 1.0
31 */
32
33 #ifndef UTILS_NEXT_COMBINATION_HPP
34 #define UTILS_NEXT_COMBINATION_HPP
35
36 #include <vector>
37 #include <iterator>
38 #include <iostream>
39 namespace oopse {
40
41 /**
42 * @brief STL next_permuationtation like combination sequence generator.
43 * Given the first and last iterator of a sequence, next_combination iteratively generates all
44 * possible combinations.
45 * @return if more combination is availiable, otherwise return false
46 * @param iterContainer iterator container
47 * @param first the first iterator
48 * @param last the last iterator
49 * @note first and last must be random access iterators and iterContainer must be the container of
50 * random access iterators . And all of the iteratos in iterContainer must be within the range [first, last)
51 *
52 * @code
53 * std::vector<int> iv;
54 * iv.push_back(1);
55 * iv.push_back(8);
56 * std::vector<std::vector<int>::iterator> ic;
57 * while(next_combination(ic, iv.begin(), iv.end())) {
58 * for (i = ic.begin(); i < ic.end(); ++i) {
59 * std::cout << **i << "\t";
60 * }
61 * std::cout << std::endl;
62 * }
63 * //output
64 * //1
65 * //8
66 * //1 8
67 * @endcode
68 */
69 template<class RandomAccessIterator, template<typename ELEM, typename = std::allocator<ELEM> > class IteratorContainer>
70 bool next_combination(IteratorContainer<RandomAccessIterator>& iterContainer, RandomAccessIterator first, RandomAccessIterator last) {
71 if (first == last) {
72 return false;
73 }
74
75 RandomAccessIterator endIter = --last;
76 typename IteratorContainer<RandomAccessIterator>::iterator i = iterContainer.end();
77
78 if (iterContainer.empty()) {
79 //if sequence is empty, we insert the first iterator
80 iterContainer.insert(iterContainer.end(), first);
81 return true;
82 } else if (*(--i) != endIter){
83 //if the last iterator in iterContainer does not reaches the end, just increase its iterator by 1
84 ++(*i);
85 return true;
86 } else {// the last iterator in iterContainer does not reaches the end
87
88 //starts at the end of the sequence and works its way towards the front, looking for two
89 //consecutive members of the sequence where the difference between them is greater
90 //than one. For example , if the sequence contains 1, 5, 8, 9 (total number is 10, first is 0
91 //and the last is 10 (due to STL's half open range)). At the end of while
92 //loop, j will point to 5, and i will point to 8, next combination should be 1, 6, 7, 8.
93 //If j is less than zero, it means it already reaches the last combination of current size.
94 //For instance, sequence may contain 6, 7, 8, 9 at this time, we need to increase the size
95 // of combination to 5
96 typename IteratorContainer<RandomAccessIterator>::iterator j = i;
97 j--;
98 while( j >= iterContainer.begin() && *i == *j + 1){
99 i--;
100 j--;
101 };
102
103 RandomAccessIterator raIter;
104 if (j - iterContainer.begin() < 0) { //reaches the last combination of current size
105 //half open range
106 if (last - first + 1 == iterContainer.size()) {
107 //if the current size equals to total number, done
108 return false;
109 } else {
110
111 //push the first currentSize+1 element into sequence
112
113 for(i = iterContainer.begin(), raIter= first; i != iterContainer.end(); ++i, ++raIter) {
114 *i = raIter;
115 }
116 iterContainer.insert(iterContainer.end(), raIter);
117
118 return true;
119 }
120
121 } else {
122 ++(*j);
123 raIter = *j;
124 for(; i != iterContainer.end(); ++i) {
125 ++raIter;
126 *i = raIter;
127 }
128 return true;
129 }
130 }
131 } //end next_combination
132
133 } //end namespace oopse
134 #endif //UTILS_NEXT_COMBINATION_HPP
135

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