本文整理汇总了C++中tree::findNodeByName方法的典型用法代码示例。如果您正苦于以下问题：C++ tree::findNodeByName方法的具体用法？C++ tree::findNodeByName怎么用？C++ tree::findNodeByName使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类tree的用法示例。
在下文中一共展示了tree::findNodeByName方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: sameTreeTolopogy
bool sameTreeTolopogy(tree t1, tree t2){
if (t1.getNodesNum() != t2.getNodesNum()) {
errorMsg::reportError("error in function same tree topology (1)");
}
tree::nodeP x = t2.getRoot();
while (x->getNumberOfSons() > 0) x= x->getSon(0);
t1.rootAt(t1.findNodeByName(x->name())->father()); // now they have the same root
t2.rootAt(t2.findNodeByName(x->name())->father()); // now they have the same root
map<int,string> names1;
treeIterDownTopConst tit1(t1);
for (tree::nodeP nodeM = tit1.first(); nodeM != tit1.end(); nodeM = tit1.next()) {
vector<string> nameOfChild;
for (int i=0; i < nodeM->getNumberOfSons();++i) {
nameOfChild.push_back(names1[nodeM->getSon(i)->id()]);
}
if (nodeM->getNumberOfSons()==0) nameOfChild.push_back(nodeM->name());
sort(nameOfChild.begin(),nameOfChild.end());
string res = "(";
for (int k=0; k < nameOfChild.size(); ++k) {
res += nameOfChild[k];
}
res += ")";
names1[nodeM->id()] = res;
}
map<int,string> names2;
treeIterDownTopConst tit2(t2);
for (tree::nodeP nodeM2 = tit2.first(); nodeM2 != tit2.end(); nodeM2 = tit2.next()) {
vector<string> nameOfChild;
for (int i=0; i < nodeM2->getNumberOfSons();++i) {
nameOfChild.push_back(names2[nodeM2->getSon(i)->id()]);
}
if (nodeM2->getNumberOfSons()==0) nameOfChild.push_back(nodeM2->name());
sort(nameOfChild.begin(),nameOfChild.end());
string res = "(";
for (int k=0; k < nameOfChild.size(); ++k) {
res += nameOfChild[k];
}
res += ")";
names2[nodeM2->id()] = res;
}
return names1[t1.getRoot()->id()] == names2[t2.getRoot()->id()];
}示例2: cutTreeToTwo
// bigTree is passed by value and not by reference. Therefore, this method doens't change the original bigTree,
// but allocates a new bigTree to be split.
bool cutTreeToTwo(tree bigTree,
const string& nameOfNodeToCut,
tree &small1,
tree &small2){// cutting above the NodeToCut.
// we want to cut the tree in two.
// first step: we make a new node between the two nodes that have to be splited,
tree::nodeP node2splitOnNewTree = bigTree.findNodeByName(nameOfNodeToCut);
string interNode = "interNode";
if (node2splitOnNewTree->father() == NULL) return(false);
// assert(node2splitOnNewTree->father() != NULL);
tree::nodeP tmp = makeNodeBetweenTwoNodes(bigTree,node2splitOnNewTree->father(),node2splitOnNewTree, interNode);
bigTree.rootAt(tmp); // tmp is the interNode and it's now the root of the tree. Its sons are node2splitOnNewTree and its father.
string allNodes = "Runs/testBifurcating/beforeCut.tree";
bigTree.output(allNodes, tree::PHYLIP, true);
cutTreeToTwoSpecial(bigTree,tmp, small1,small2);
if (small1.getNodesNum() < 5 || small2.getNodesNum() < 5) return (false);
LOGDO(15,small1.output(myLog::LogFile(),tree::ANCESTORID));
LOGDO(15,small2.output(myLog::LogFile(),tree::ANCESTORID));
tree::nodeP toDel1 = small1.findNodeByName(interNode);
small1.removeLeaf(toDel1);
tree::nodeP toDel2 = small2.findNodeByName(interNode);
small2.removeLeaf(toDel2);
// this part fix the ids.
treeIterTopDown tIt(small1);
int newId =0;
for (tree::nodeP mynode = tIt.first(); mynode != tIt.end(); mynode = tIt.next()) {
mynode->setID(newId);
newId++;
}
treeIterTopDown tIt2(small2);
int newId2 =0;
for (tree::nodeP mynode2 = tIt2.first(); mynode2 != tIt2.end(); mynode2 = tIt2.next()) {
mynode2->setID(newId2);
newId2++;
}
return (true); // successes!
};