A *算法：关闭列表包含太多元素/太大

Question

我目前正在用JavaScript实现A *算法。但是，我遇到了一个问题：我的closedList似乎太大了。以下是输出的屏幕截图：

什么可能导致这个问题？我的启发式计算错了吗？

Node.prototype.getHeuristic = function(pos0, pos1)
{
    // Manhatten Distance
    var horizontalDistance = Math.abs(pos1.x - pos0.x);
    var verticalDistance = Math.abs(pos1.y - pos0.y);
    return horizontalDistance + verticalDistance;
}

或者我在这个方法中理解/实现了什么错误？：

PathFinder.prototype.findPath = function() 
{
var start = new Date().getTime();
var openList = [];
var closedList = [];

var startNode = this.startNode;
var grid = this.grid;
var endNode = this.finishNode;



openList.push(startNode);

while (openList.length > 0)
{
    var lowInd = 0;
    for(var i = 0; i < openList.length; i++) {
        if (openList[i].f < openList[lowInd].f) 
        {
            lowInd = i; 
        }
    }
    var currentNode = openList[lowInd];



    if (currentNode.x == endNode.x && currentNode.y == endNode.y)
    {
        var curr = currentNode;
        var ret = [];
        while (curr.parent)
        {
            ret.push(curr);
            curr.type = NODES.PATH;
            curr = curr.parent;
        }   

        this.finishNode.type = NODES.FINISH;
        this.printGrid();   
        console.log("Total Operations: " + this.operations);

        var end = new Date().getTime();
        var time = end - start;
        console.log('Execution time: ' + time + "ms");

        return true;
    }


    openList.splice(lowInd, 1);
    closedList.push(currentNode);
    if (currentNode.type != NODES.START) 
    {
        currentNode.type = NODES.CLOSED;
    }

    var neighbors = currentNode.getNeighbors(this.grid);

 for (var indexNeighbors = 0; indexNeighbors < neighbors.length; indexNeighbors++)
    {
        var neighbor = neighbors[indexNeighbors];

        if (this.findNodeInArray(closedList, neighbor) || neighbor.isWall())
        {
            continue;
        }

        var gValue = currentNode.g + 1;
        var isGvalueLowest = false;

        if (!this.findNodeInArray(openList, neighbor))
        {
            isGvalueLowest = true;
            neighbor.h = neighbor.getHeuristic(neighbor, endNode);
            openList.push(neighbor);
        } 
        else if (gValue < neighbor.g) 
        {
            isGvalueLowest = true;
        }

        if (isGvalueLowest) 
        {
            neighbor.parent = currentNode;
            neighbor.g = gValue;
            neighbor.f = neighbor.g + neighbor.h;   
            neighbor.type = NODES.MARKED;

            console.log(neighbor);
            this.operations++;
        }
    }

}
}

如果您想查看代码的更多部分，请告诉我。感谢您的帮助，谢谢。

Answer 1

您需要break ties towards the endpoint。

（不破坏与端点的联系）

（与端点断开关系）

（有障碍物的例子）