在STD中崩溃:: Sort-分类而没有严格的弱排序

我正在尝试对项目进行分类。如代码注释中所述，订购应为：

具有更多动作点(mAp(的参与者首先。当有领带时，与战斗的发起人(mBattleInitiator(的参与者(mDisposition(首先。

以下代码(简化示例(在MacOS上崩溃，大概是由于我的排序实现不正确：

#include <QtCore>
class AiComponent
{
public:
    enum Disposition {
        Friendly,
        Hostile
    };
    AiComponent(Disposition disposition) : mDisposition(disposition) {}
    ~AiComponent() { qDebug() << "Destroying AiComponent"; }
    Disposition mDisposition;
};
class BattleManager
{
public:
    BattleManager() : mBattleInitiator(AiComponent::Hostile) {}
    class Turn {
    public:
        Turn() : mAp(1) {}
        Turn(QSharedPointer<AiComponent> aiComponent) :
            mAiComponent(aiComponent),
            mAp(1)
        {
        }
        Turn(const Turn &rhs) :
            mAiComponent(rhs.mAiComponent),
            mAp(1)
        {
        }
        QSharedPointer<AiComponent> mAiComponent;
        int mAp;
    };
    void addToTurnQueue(QSet<QSharedPointer<AiComponent>> aiComponents);
    AiComponent::Disposition mBattleInitiator;
    QVector<Turn> mTurnQueue;
    Turn mActive;
};
void BattleManager::addToTurnQueue(QSet<QSharedPointer<AiComponent> > aiComponents)
{
    foreach (auto aiComponent, aiComponents) {
        mTurnQueue.append(Turn(aiComponent));
    }
    // Sort the participants so that ones with more action points (mAp) go first.
    // When there is a tie, participants with the same disposition (mDisposition)
    // as the initiator of the battle (mBattleInitiator) go first.
    std::sort(mTurnQueue.begin(), mTurnQueue.end(), [=](const Turn &a, const Turn &b) {
        if (a.mAp > b.mAp)
            return true;
        if (a.mAp < b.mAp)
            return false;
        // At this point, a.mAp is equal to b.mAp, so we must resolve the tie
        // based on mDisposition.
        if (a.mAiComponent->mDisposition == mBattleInitiator)
            return true;
        if (b.mAiComponent->mDisposition == mBattleInitiator)
            return false;
        return false;
    });
}
int main(int /*argc*/, char */*argv*/[])
{
    BattleManager battleManager;
    for (int i = 0; i < 20; ++i) {
        qDebug() << "iteration" << i;
        QSet<QSharedPointer<AiComponent>> participants;
        AiComponent::Disposition disposition = i % 2 == 0 ? AiComponent::Hostile : AiComponent::Friendly;
        QSharedPointer<AiComponent> ai(new AiComponent(disposition));
        participants.insert(ai);
        battleManager.addToTurnQueue(participants);
    }
    // This should print (1 1), (1 1), ... (1 0), (1 0)
    foreach (auto turn, battleManager.mTurnQueue) {
        qDebug() << "(" << turn.mAp << turn.mAiComponent->mDisposition << ")";
    }
    return 0;
}

我已经研究了有关该主题的其他答案。他们中的大多数只是说"将其实施为a> b"，就我而言，这将无法使用。有一些看起来很重要，但没有帮助我：

• https://stackoverflow.com/a/16824720/904422-说其他标准算法将起作用，但没有提及任何具体示例
• https://stackoverflow.com/a/33508373/904422-令人困惑，看起来像Overkill

最简单的实现我追求的方法是什么？