本文整理汇总了C++中CvCity::unhappyLevel方法的典型用法代码示例。如果您正苦于以下问题：C++ CvCity::unhappyLevel方法的具体用法？C++ CvCity::unhappyLevel怎么用？C++ CvCity::unhappyLevel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CvCity的用法示例。
在下文中一共展示了CvCity::unhappyLevel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: operator
result_type operator() (const ResourceInfo::BaseNode& node)
{
info.rawHappy = node.baseHappy;
for (size_t i = 0, count = node.buildingNodes.size(); i < count; ++i)
{
(*this)(node.buildingNodes[i]);
}
CityIter iter(CvPlayerAI::getPlayer(playerType_));
CvCity* pLoopCity;
// how many citizens are made happy now if we had the resource, how many would be happy if we had the requisite buildings,
// and how many more could be made happy with the buildings we have.
// E.g. suppose we have 1 unhappy citizen in a city, with a forge - and we analyse gold - this would give:
// actualHappy = 1, potentialHappy = 0, unusedHappy = 1, but if the city had no forge, it should give:
// actualHappy = 1, potentialHappy = 1, unusedHappy = 0
int actualHappy = 0, potentialHappy = 0, unusedHappy = 0;
while (pLoopCity = iter())
{
int unhappyCitizens = std::max<int>(0, pLoopCity->unhappyLevel() - pLoopCity->happyLevel());
int baseHappy = std::min<int>(node.baseHappy, unhappyCitizens);
// have we got any left?
unhappyCitizens = std::max<int>(0, unhappyCitizens - baseHappy);
if (unhappyCitizens == 0)
{
unusedHappy += node.baseHappy - baseHappy;
}
actualHappy += baseHappy;
for (size_t i = 0, count = info.buildingHappy.size(); i < count; ++i)
{
int buildingCount = pLoopCity->getNumBuilding(info.buildingHappy[i].first);
if (buildingCount > 0)
{
int thisBuildingCount = std::min<int>(buildingCount * info.buildingHappy[i].second, unhappyCitizens);
unhappyCitizens = std::max<int>(0, unhappyCitizens - thisBuildingCount);
if (unhappyCitizens == 0)
{
unusedHappy += buildingCount * info.buildingHappy[i].second - thisBuildingCount;
}
actualHappy += thisBuildingCount;
}
else // just assumes one building allowed (should use getCITY_MAX_NUM_BUILDINGS(), but this would be wrong if building is limited in some other way, e.g. wonders)
{
potentialHappy += info.buildingHappy[i].second;
}
}
}
info.actualHappy = actualHappy;
info.potentialHappy = potentialHappy;
info.unusedHappy = unusedHappy;
}