本文整理汇总了C++中StateSet::getStates方法的典型用法代码示例。如果您正苦于以下问题：C++ StateSet::getStates方法的具体用法？C++ StateSet::getStates怎么用？C++ StateSet::getStates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StateSet的用法示例。
在下文中一共展示了StateSet::getStates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:
StateTreeNode::StateTreeNode(
const StateSet &stateSet,
int maxDepth,
double centerShiftMultiplier
) :
numSpacePartitions(pow(2, stateSet.getStates().at(0)->getCoordinates().size()))
{
const vector<AbstractState*> &states = stateSet.getStates();
unsigned int numCoordinates = states.at(0)->getCoordinates().size();
for(unsigned int n = 1; n < states.size(); n++){
TBTKAssert(
numCoordinates == states.at(n)->getCoordinates().size(),
"StateTreeNode::StateTreeNode()",
"Unable to handle StateSets containing states with different dimensions.",
""
);
}
//Find bounding box.
vector<double> min;
vector<double> max;
double maxFiniteExtent = 0.;
unsigned int n = 0;
//Find first state with finite extent and initialize min and max such
//that the state is contained inside the corresponding rectangle.
for(; n < states.size(); n++){
if(states.at(n)->hasFiniteExtent()){
for(unsigned int c = 0; c < numCoordinates; c++){
min.push_back(states.at(n)->getCoordinates().at(c) - states.at(n)->getExtent());
max.push_back(states.at(n)->getCoordinates().at(c) + states.at(n)->getExtent());
}
maxFiniteExtent = states.at(n)->getExtent();
break;
}
}
//If all state have infinite extent, create a dummy box of zero extent.
if(n == states.size()){
for(unsigned int c = 0; c < states.at(0)->getCoordinates().size(); c++){
min.push_back(0);
max.push_back(0);
}
}
//Loop over all remaining states and update min and max so that every
//state is contained in the corresponding rectangle.
for(; n < states.size(); n++){
if(!states.at(n)->hasFiniteExtent())
continue;
for(unsigned int c = 0; c < numCoordinates; c++){
if(min.at(c) > states.at(n)->getCoordinates().at(c) - states.at(n)->getExtent())
min.at(c) = states.at(n)->getCoordinates().at(c) - states.at(n)->getExtent();
if(max.at(c) < states.at(n)->getCoordinates().at(c) + states.at(n)->getExtent())
max.at(c) = states.at(n)->getCoordinates().at(c) + states.at(n)->getExtent();
}
if(maxFiniteExtent < states.at(n)->getExtent())
maxFiniteExtent = states.at(n)->getExtent();
}
//Enlarge and shift box to avoid problems with high level partition
//boundaries cutting through states. This would be particularly
//problematic if the states for example have three dimensional
//coordinates, but are contained in a plane. Without shifting the box,
//the first partition boundary would cut every state and thereby every
//state would be added to the root node.
for(unsigned int n = 0; n < max.size(); n++)
max.at(n) += centerShiftMultiplier*maxFiniteExtent;
for(unsigned int n = 0; n < min.size(); n++)
min.at(n) -= 2.*(max.at(n) - min.at(n))*(1 - ROUNDOFF_MARGIN_MULTIPLIER);
//Calculate center nad halfSize of the bounding box.
halfSize = 0.;
for(unsigned int n = 0; n < numCoordinates; n++){
center.push_back((min.at(n) + max.at(n))/2.);
if(halfSize < (max.at(n) - min.at(n))/2.)
halfSize = (max.at(n) - min.at(n))/2.;
}
this->maxDepth = maxDepth;
for(unsigned int n = 0; n < states.size(); n++)
add(states.at(n));
}