Dialogue: improve conflict resolution for chains of overlapping keywords (unit test for such a case will follow)

openmw-35
scrawl 10 years ago
parent 65ab31eae6
commit ce2cbab402

@ -66,8 +66,14 @@ public:
return false;
}
static bool sortMatches(const Match& left, const Match& right)
{
return left.mBeg < right.mBeg;
}
void highlightKeywords (Point beg, Point end, std::vector<Match>& out)
{
std::vector<Match> matches;
for (Point i = beg; i != end; ++i)
{
// check if previous character marked start of new word
@ -144,42 +150,51 @@ public:
match.mValue = candidate->second.mValue;
match.mBeg = i;
match.mEnd = k;
out.push_back(match);
matches.push_back(match);
break;
}
}
// resolve overlapping keywords
for (typename std::vector<Match>::iterator it = out.begin(); it != out.end();)
while (matches.size())
{
typename std::vector<Match>::iterator next = it;
++next;
if (next == out.end())
break;
if (it->mEnd <= next->mBeg)
{
++it;
continue; // no overlap
}
else
int longestKeywordSize = 0;
typename std::vector<Match>::iterator longestKeyword;
for (typename std::vector<Match>::iterator it = matches.begin(); it != matches.end(); ++it)
{
// prefer the longer keyword
int size = it->mEnd - it->mBeg;
int nextSize = next->mEnd - next->mBeg;
if (size >= nextSize) // if both are the same length, then prefer the first keyword
if (size > longestKeywordSize)
{
out.erase(next);
continue;
longestKeywordSize = size;
longestKeyword = it;
}
else
typename std::vector<Match>::iterator next = it;
++next;
if (next == matches.end())
break;
if (it->mEnd <= next->mBeg)
{
it = out.erase(it);
continue;
break; // no overlap
}
}
Match keyword = *longestKeyword;
matches.erase(longestKeyword);
out.push_back(keyword);
// erase anything that overlaps with the keyword we just added to the output
for (typename std::vector<Match>::iterator it = matches.begin(); it != matches.end();)
{
if (it->mBeg < keyword.mEnd && it->mEnd > keyword.mBeg)
it = matches.erase(it);
else
++it;
}
}
std::sort(out.begin(), out.end(), sortMatches);
}
private:

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