Question

有没有更好的方法来处理我的A Star算法的FindAdjacent（）函数？它非常混乱，并且它没有正确设置父节点。当它试图找到路径时，它会无限循环，因为节点的父节点有一个节点的五边形，而父节点总是彼此。

任何帮助都会很棒。这是我的功能：

void AStarImpl::FindAdjacent(Node* pNode)
{
    for (int i = -1; i <= 1; i++)
    {
        for (int j = -1; j <= 1; j++)
        {
            if (pNode->mX != Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mX 
                || pNode->mY != Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mY)
            {
                if (pNode->mX + i <= 14 && pNode->mY + j <= 14)
                {
                    if (pNode->mX + i >= 0 && pNode->mY + j >= 0)
                    {
                        if (Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mTypeID != NODE_TYPE_SOLID)
                        {
                            if (find(mOpenList.begin(), mOpenList.end(), &Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j]) == mOpenList.end())
                            {

                                Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j].mParent = &Map::GetInstance()->mMap[pNode->mX][pNode->mY];
                                mOpenList.push_back(&Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j]);
                            }
                        }
                    }
                }

            }
        }
    }

    mClosedList.push_back(&Map::GetInstance()->mMap[pNode->mX][pNode->mY]);
}

如果您想要更多代码，请询问，我可以发布。

Answer 1

您可以使用continue减少嵌套ifs的数量。一般来说，以下两个代码块是等效的：

while(conditionA){
    if(conditionB){
        doStuff();
    }
}

while(conditionA){
    if (!conditionB){continue;}
    doStuff();
}

我们可以使用此原则来减少代码中嵌套ifs的数量。

void AStarImpl::FindAdjacent(Node* pNode)
{
    for (int i = -1; i <= 1; i++)
    {
        for (int j = -1; j <= 1; j++)
        {
            if (pNode->mX == Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mX && pNode->mY == Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mY){continue;}
            if (pNode->mX + i > 14 || pNode->mY + j > 14){continue;}
            if (pNode->mX + i < 0 || pNode->mY + j < 0){continue;}
            if (Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mTypeID == NODE_TYPE_SOLID){continue;}
            if (find(mOpenList.begin(), mOpenList.end(), &Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j]) != mOpenList.end()){continue;}
            Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j].mParent = &Map::GetInstance()->mMap[pNode->mX][pNode->mY];
            mOpenList.push_back(&Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j]);
        }
    }

    mClosedList.push_back(&Map::GetInstance()->mMap[pNode->mX][pNode->mY]);
}

如果我正确理解了您的第一个if条件，那么您只是试图断言pNode不是它自己的邻居。在这种情况下，您可以将代码更改为：

void AStarImpl::FindAdjacent(Node* pNode)
{
    for (int i = -1; i <= 1; i++)
    {
        for (int j = -1; j <= 1; j++)
        {
            if (i == 0 && j == 0){continue;}
            if (pNode->mX + i > 14 || pNode->mY + j > 14){continue;}
            if (pNode->mX + i < 0 || pNode->mY + j < 0){continue;}
            if (Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mTypeID == NODE_TYPE_SOLID){continue;}
            if (find(mOpenList.begin(), mOpenList.end(), &Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j]) != mOpenList.end()){continue;}
            Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j].mParent = &Map::GetInstance()->mMap[pNode->mX][pNode->mY];
            mOpenList.push_back(&Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j]);
        }
    }

    mClosedList.push_back(&Map::GetInstance()->mMap[pNode->mX][pNode->mY]);
}

理想情况下，您的FindAdjacent方法根本不必修改打开或关闭集。相反，让它返回所有邻居，无论它们是打开还是关闭。如果要将这些邻居添加到打开或关闭的集合中，或者检查它们是否是这些集合的成员，那么应该在实际实现aStar算法的方法中完成。

Vector<Node> AStarImpl:FindAdjacent(Node* pNode)
{
    Vector<Node> neighbors;
    for (int i = -1; i <= 1; i++)
    {
        for (int j = -1; j <= 1; j++)
        {
            if (i == 0 && j == 0){continue;}
            if (pNode->mX + i > 14 || pNode->mY + j > 14){continue;}
            if (pNode->mX + i < 0 || pNode->mY + j < 0){continue;}
            if (Map::GetInstance()->mMap[pNode->mX + i][pNode->mY + j].mTypeID == NODE_TYPE_SOLID){continue;}
            neighbors.push_back(Map::GetInstance()->mMap[pNode->mX+i][pNode->mY+j]);
        }
    }
    return neighbors;
}

您多次执行一些相同的操作。通过将这些操作的结果存储在变量中，您可以更清楚地表达您的意图。这样做不会使您的代码更短，但可能会使其更具可读性。

Vector<Node> AStarImpl:FindAdjacent(Node* pNode)
{
    Vector<Node> neighbors;
    for (int i = -1; i <= 1; i++)
    {
        for (int j = -1; j <= 1; j++)
        {
            if (i == 0 && j == 0){continue;}
            int x = pNode->mX + i;
            int y = pNode->mY + j;
            if (x > 14 || y > 14){continue;}
            if (x < 0 || y < 0){continue;}
            Node candidate = Map::GetInstance()->mMap[x][y];
            if (candidate.mTypeID == NODE_TYPE_SOLID){continue;}
            neighbors.push_back(candidate);
        }
    }
    return neighbors;
}

查找相邻节点星路径查找C ++

1 个答案: