本文整理汇总了C++中GoBoard::NumNeighbors方法的典型用法代码示例。如果您正苦于以下问题：C++ GoBoard::NumNeighbors方法的具体用法？C++ GoBoard::NumNeighbors怎么用？C++ GoBoard::NumNeighbors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GoBoard的用法示例。
在下文中一共展示了GoBoard::NumNeighbors方法的3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: NumberOfMoveToEye
bool GoEyeUtil::NumberOfMoveToEye(const GoBoard& board, SgBlackWhite color,
SgPoint p, int& number)
{
SG_ASSERT(board.IsEmpty(p));
SgBlackWhite att = SgOppBW(color); // attacker
if ( board.Line(p) == 1) // corner or edge
{
if ( board.Num8Neighbors(p, att) > 0 )
return false;
else
{
number = board.Num8EmptyNeighbors(p);
return true;
}
}
else // in center
{
if ( board.Num8Neighbors(p, att) >= 2 )
return false;
else if ( board.NumNeighbors(p, att) > 0 )
return false;
else // only 0 or 1 attacker point and not in NB4
{
number = board.Num8EmptyNeighbors(p);
return true;
}
}
}示例2: IsVitalPt
bool GoEyeUtil::IsVitalPt(const SgPointSet& points, SgPoint p,
SgBlackWhite opp,
const GoBoard& bd)
{
// in corridors a vital point has 2 nbs, in big ones it may have 3 or 4.
// but: 2 in following
// example: unsettled nakade, non-flat points are all occupied by opp.
// .a a is vital Point.
// o.
// .
int numNb = bd.NumEmptyNeighbors(p) + bd.NumNeighbors(p, opp);
if (numNb >= 2)
{
if (numNb >= 4)
/* */ return true; /* */
int nu = IsTreeShape(points) ? 2 : 3;
if (numNb >= nu)
{
if (numNb == 2 && bd.IsEmpty(p))
return IsSplitPt(p, points);
else
return true;
}
}
return false;
}示例3: IsPossibleEye
bool GoEyeUtil::IsPossibleEye(const GoBoard& board, SgBlackWhite color,
SgPoint p)
{
bool isPossibleEye = false;
SG_ASSERT(board.GetColor(p) != color);
const SgBlackWhite opp = SgOppBW(color);
if (board.Line(p) == 1) // corner or edge
{
const int nuOwn = (board.Pos(p) == 1) ? 2 : 4;
if ( board.Num8Neighbors(p, color) == nuOwn
&& board.Num8EmptyNeighbors(p) == 1
)
{
isPossibleEye = true;
}
}
else // in center
{
// have all neighbors, and 2 diagonals, and can get a third
if ( board.NumNeighbors(p, color) == 4
&& board.NumDiagonals(p, color) == 2
&& board.NumEmptyDiagonals(p) > 0
)
{
isPossibleEye = true;
}
// have 3 of 4 neighbors, can get the 4th, and have enough diagonals
else if ( board.NumNeighbors(p, color) == 3
&& board.NumNeighbors(p, opp) == 0
&& board.NumDiagonals(p, color) >= 3
)
{
isPossibleEye = true;
}
}
return isPossibleEye;
}