本文整理汇总了C++中ContigPaths::erase方法的典型用法代码示例。如果您正苦于以下问题：C++ ContigPaths::erase方法的具体用法？C++ ContigPaths::erase怎么用？C++ ContigPaths::erase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ContigPaths的用法示例。
在下文中一共展示了ContigPaths::erase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: handleEstimate
//.........这里部分代码省略.........
= distanceMap.find(v);
if (dmIter == distanceMap.end()) {
// This contig is a repeat.
ignoredCount++;
vout << "ignored\n";
continue;
}
// translate distance by -overlap to match
// coordinate space used by the estimate
int actualDistance = dmIter->second;
int diff = actualDistance - ep.distance;
unsigned buffer = allowedError(ep.stdDev);
bool invalid = (unsigned)abs(diff) > buffer;
bool repeat = repeats.count(v.contigIndex()) > 0;
bool ignored = invalid && repeat;
if (ignored)
ignoredCount++;
else if (invalid)
invalidCount++;
else
validCount++;
vout << "dist: " << actualDistance
<< " diff: " << diff
<< " buffer: " << buffer
<< " n: " << ep.numPairs
<< (ignored ? " ignored" : invalid ? " invalid" : "")
<< '\n';
}
if (invalidCount == 0 && validCount > 0)
++solIter;
else
solIter = solutions.erase(solIter);
}
vout << "Solutions: " << solutions.size();
if (tooComplex)
vout << " (too complex)";
if (tooManySolutions)
vout << " (too many solutions)";
vout << '\n';
ContigPaths::iterator bestSol = solutions.end();
int minDiff = 999999;
for (ContigPaths::iterator solIter = solutions.begin();
solIter != solutions.end(); ++solIter) {
map<ContigNode, int> distanceMap
= makeDistanceMap(g, origin, *solIter);
int sumDiff = 0;
for (Estimates::const_iterator iter
= er.estimates[dirIdx].begin();
iter != er.estimates[dirIdx].end(); ++iter) {
ContigNode v = iter->first;
const DistanceEst& ep = iter->second;
if (repeats.count(v.contigIndex()) > 0)
continue;
map<ContigNode, int>::iterator dmIter
= distanceMap.find(v);
assert(dmIter != distanceMap.end());
int actualDistance = dmIter->second;
int diff = actualDistance - ep.distance;
sumDiff += abs(diff);
}
if (sumDiff < minDiff) {