Changeset 1578:1d3a1bcbc874 in lemon0.x for doc
 Timestamp:
 07/21/05 00:36:37 (16 years ago)
 Branch:
 default
 Phase:
 public
 Convert:
 svn:c9d7d8f590d60310b91f818b3a526b0e/lemon/trunk@2079
 File:

 1 edited
Legend:
 Unmodified
 Added
 Removed

doc/quicktour.dox
r1541 r1578 142 142 143 143 144 <li> If you want to design a network and want to minimize the total length 145 of wires then you might be looking for a <b>minimum spanning tree</b> in 146 an undirected graph. This can be found using the Kruskal algorithm: the 147 function \ref lemon::kruskal "LEMON Kruskal ..." does this job for you. 148 The following code fragment shows an example: 149 150 Ide Zsuzska fog irni! 144 <li> If you want to design a network and want to minimize the total 145 length of wires then you might be looking for a <b>minimum spanning 146 tree</b> in an undirected graph. This can be found using the Kruskal 147 algorithm: the function \ref lemon::kruskal "LEMON Kruskal " does 148 this job for you. After we had a graph \c g and a cost map \c 149 edge_cost_map , the following code fragment shows an example how to get weight of the minmum spanning tree, if the costs are uniform: 150 151 \dontinclude kruskal_demo.cc 152 \skip std::cout 153 \until kruskal 154 155 It gives back a edge bool map, which contains the edges of the tree. 156 If the costs are nonuniform, for example the cost is given by \c 157 edge_cost_map_2 , or the edges of the tree are have to be given in a 158 vector, then we can give to the kruskal a vector \c tree_edge_vec , instead of 159 an edge bool map: 160 161 \skip edge_cost_map_2 162 \until edge_cost_map_2, std::back_inserter 163 164 And finally the next fragment shows how to use the functions \c makeKruskalMapInput and \c makeKruskalSequenceOutPut: 165 166 \skip makeKruskalSequenceOutput 167 \until tree_edge_vec 168 169 See the whole program in \ref kruskal_demo.cc. 170 171 151 172 152 173 <li>Many problems in network optimization can be formalized by means
Note: See TracChangeset
for help on using the changeset viewer.