Unity A*寻径崩溃

本文关键字:崩溃 Unity | 更新日期: 2023-09-27 17:54:31

所以我试图实现A*算法。基本逻辑基本完成了,但有一个问题我就是无法解决。

当请求路径时,Unity停止响应,我的PC一直变慢,直到它挂起,我不得不强制关闭。

下面是简化后的代码:
public static List<Node> ReturnPath(Vector3 pointA, Vector3 pointB)
{
    /* A Bunch of initializations */
    while (!pathFound) 
    {
        //add walkable neighbours to openlist
        foreach (Node n in GetNeighbours(currentNode)) 
        {
            if (!n.walkable)
                continue;
            n.gCost = currentNode.gCost + GetManDist (currentNode, n);
            n.hCost = GetManDist (n, B);
            openList.Add (n);
            n.parent = currentNode;
        }
        //check openList for lowest fcost or lower hCost
        for (int i = 0; i < openList.Count; i++) 
        {
            if ((openList [i].fCost < currentNode.fCost) || (openList [i].fCost == currentNode.fCost && openList [i].hCost < currentNode.hCost)) 
            {
                //make the currentNode = node with lowest fcost
                currentNode = openList [i];
            }
            openList.Remove (currentNode);
            if(!closedList.Contains(currentNode))
                closedList.Add (currentNode);
        }
        //Check if the currentNode it destination
        if (currentNode == B) 
        {
            Debug.Log ("Path Detected");
            path = RetracePath (A, B);
            pathFound = true;
        }
    }
    return path;
}

只要目的地在两个节点之外,就可以正常工作。如果超过这个数,就会出现上述问题。为什么呢?我的第一个猜测是我在openList中投入了太多。

注意:我将一个30 × 30的单位平台(层)分成1x1的正方形,称为"节点"。GetManDist()获取两个节点之间的曼哈顿距离。

UPDATE:这是工作代码。注释太长

    public List<Node> ReturnPath(Vector3 pointA, Vector3 pointB)
{
    List<Node> openList = new List<Node>();
    List<Node> closedList = new List<Node>();
    List<Node> path = new List<Node> ();
    Node A = ToNode (pointA);
    Node B = ToNode (pointB);
    Node currentNode;
    bool pathFound = false;
    A.hCost = GetManDist(A, B);
    A.gCost = 0;
    openList.Add (A);
    while (!pathFound) // while(openList.Count > 0) might be better because it handles the possibility of a non existant path
    {
        //Set to arbitrary element
        currentNode = openList [0];
        //Check openList for lowest fcost or lower hCost
        for (int i = 0; i < openList.Count; i++) 
        {
            if ((openList [i].fCost < currentNode.fCost) || ((openList [i].fCost == currentNode.fCost && openList [i].hCost < currentNode.hCost))) 
            {
                //Make the currentNode = node with lowest fcost
                currentNode = openList [i];
            }
        }
        //Check if the currentNode is destination
        if (currentNode.hCost == 0) //Better than if(currentNode == B)
        {
            path = RetracePath (A, B);
            pathFound = true;
        }
        //Add walkable neighbours to openlist
        foreach (Node n in GetNeighbours(currentNode)) 
        {
            //Avoid wasting time checking unwalkable and those already checked
            if (!n.walkable || closedList.Contains(n))
                continue;
            //Movement Cost to neighbour
            int newGCost = currentNode.gCost + GetManDist (currentNode, n);
            //Calculate g_Cost, Update if new g_cost to neighbour is less than an already calculated one
            if (n.gCost > newGCost || !openList.Contains (n)) 
            {
                n.gCost = newGCost;
                n.hCost = GetManDist (n, B);
                n.parent = currentNode; //So we can retrace
                openList.Add (n);
            }
        }
        //We don't need you no more...
        openList.Remove (currentNode);
        //Avoid redundancy of nodes in closedList
        if(!closedList.Contains(currentNode))
            closedList.Add (currentNode);
    }
    return path;
}

Unity A*寻径崩溃

currentNode的值有问题。由于我们在openlist中检查具有最低f_Cost或较低h_Cost的节点,而不是currentNode,因此在某些情况下,当寻路遇到墙时,它必须返回或转向导致增加f_Cost和h_Cost(两者都大于currentNode)。因此,openlist中不再有f_Cost/h_Cost较低的节点导致无限循环。简单的解决方案是每次将currentNode设置为openList中的任意元素。

添加

currentNode = openlist[0];