本文整理汇总了C++中HoleCollection::Add方法的典型用法代码示例。如果您正苦于以下问题：C++ HoleCollection::Add方法的具体用法？C++ HoleCollection::Add怎么用？C++ HoleCollection::Add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HoleCollection的用法示例。
在下文中一共展示了HoleCollection::Add方法的1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: addHieroRule
// this function is called recursively
// it pokes a new hole into the phrase pair, and then calls itself for more holes
void ExtractTask::addHieroRule( int startT, int endT, int startS, int endS
, RuleExist &ruleExist, HoleCollection &holeColl
, int numHoles, int initStartT, int wordCountT, int wordCountS)
{
// done, if already the maximum number of non-terminals in phrase pair
if (numHoles >= m_options.maxNonTerm)
return;
// find a hole...
for (int startHoleT = initStartT; startHoleT <= endT; ++startHoleT) {
for (int endHoleT = startHoleT+(m_options.minHoleTarget-1); endHoleT <= endT; ++endHoleT) {
// if last non-terminal, enforce word count limit
if (numHoles == m_options.maxNonTerm-1 && wordCountT - (endHoleT-startT+1) + (numHoles+1) > m_options.maxSymbolsTarget)
continue;
// determine the number of remaining target words
const int newWordCountT = wordCountT - (endHoleT-startHoleT+1);
// always enforce min word count limit
if (newWordCountT < m_options.minWords)
continue;
// except the whole span
if (startHoleT == startT && endHoleT == endT)
continue;
// does a phrase cover this target span?
// if it does, then there should be a list of mapped source phrases
// (multiple possible due to unaligned words)
const HoleList &sourceHoles = ruleExist.GetSourceHoles(startHoleT, endHoleT);
// loop over sub phrase pairs
HoleList::const_iterator iterSourceHoles;
for (iterSourceHoles = sourceHoles.begin(); iterSourceHoles != sourceHoles.end(); ++iterSourceHoles) {
const Hole &sourceHole = *iterSourceHoles;
const int sourceHoleSize = sourceHole.GetEnd(0)-sourceHole.GetStart(0)+1;
// enforce minimum hole size
if (sourceHoleSize < m_options.minHoleSource)
continue;
// determine the number of remaining source words
const int newWordCountS = wordCountS - sourceHoleSize;
// if last non-terminal, enforce word count limit
if (numHoles == m_options.maxNonTerm-1 && newWordCountS + (numHoles+1) > m_options.maxSymbolsSource)
continue;
// enforce min word count limit
if (newWordCountS < m_options.minWords)
continue;
// hole must be subphrase of the source phrase
// (may be violated if subphrase contains additional unaligned source word)
if (startS > sourceHole.GetStart(0) || endS < sourceHole.GetEnd(0))
continue;
// make sure target side does not overlap with another hole
if (holeColl.OverlapSource(sourceHole))
continue;
// if consecutive non-terminals are not allowed, also check for source
if (!m_options.nonTermConsecSource && holeColl.ConsecSource(sourceHole) )
continue;
// check that rule scope would not exceed limit if sourceHole
// were added
if (holeColl.Scope(sourceHole) > m_options.maxScope)
continue;
// require that at least one aligned word is left (unless there are no words at all)
if (m_options.requireAlignedWord && (newWordCountS > 0 || newWordCountT > 0)) {
HoleList::const_iterator iterHoleList = holeColl.GetHoles().begin();
bool foundAlignedWord = false;
// loop through all word positions
for(int pos = startT; pos <= endT && !foundAlignedWord; pos++) {
// new hole? moving on...
if (pos == startHoleT) {
pos = endHoleT;
}
// covered by hole? moving on...
else if (iterHoleList != holeColl.GetHoles().end() && iterHoleList->GetStart(1) == pos) {
pos = iterHoleList->GetEnd(1);
++iterHoleList;
}
// covered by word? check if it is aligned
else {
if (m_sentence.alignedToT[pos].size() > 0)
foundAlignedWord = true;
}
}
if (!foundAlignedWord)
continue;
}
// update list of holes in this phrase pair
holeColl.Add(startHoleT, endHoleT, sourceHole.GetStart(0), sourceHole.GetEnd(0));
//.........这里部分代码省略.........