A* 算法最快时间

Question

我已经实现了 A* 算法来给出最短距离的路线，但是我正在尝试改变它，以便它可以计算出最快的路线。 使用这个伪代码：

function A*(start,goal)
    closedset := the empty set    // The set of nodes already evaluated.
    openset := {start}    // The set of tentative nodes to be evaluated, initially containing the start node
    came_from := the empty map    // The map of navigated nodes.

    g_score[start] := 0    // Cost from start along best known path.
    // Estimated total cost from start to goal through y.
    f_score[start] := g_score[start] + heuristic_cost_estimate(start, goal)

    while openset is not empty
        current := the node in openset having the lowest f_score[] value
        if current = goal
            return reconstruct_path(came_from, goal)

        remove current from openset
        add current to closedset
        for each neighbor in neighbor_nodes(current)
            if neighbor in closedset
                continue
            tentative_g_score := g_score[current] + dist_between(current,neighbor)

            if neighbor not in openset or tentative_g_score < g_score[neighbor] 
                came_from[neighbor] := current
                g_score[neighbor] := tentative_g_score
                f_score[neighbor] := g_score[neighbor] + heuristic_cost_estimate(neighbor, goal)
                if neighbor not in openset
                    add neighbor to openset

    return failure

我认为计算最快路线的最简单方法是将当前和邻居之间的距离除以该道路的限速：tentative_g_score := g_score[current] + (dist_between(current,neighbor)/neighbor.速度极限） 但是，这并没有在我的算法中给出正确的结果。

谁能指出我正确的方向如何有效地做到这一点？

这是我当前的代码：http://pastebin.com/QWi6AwF9

最快的路线，即从起点到达目的地所需时间最短的路线。

我的启发式函数是这样的

private double heuristic(Vertex goal, Vertex next) 
    {
        return (Math.sqrt(Math.pow((goal.x - next.x), 2) + Math.pow((goal.y - next.y), 2)));
    }

干杯

Answer 1

启发式函数必须是可接受的（也就是说，它不应该高估到目标的距离）。一旦你开始将边的长度除以speedlimit，到目标的实际距离可能小于到它的欧几里得距离（如果speedlimit > 1）。它破坏了算法。如何解决？例如，您可以使用dist / MAX_SPEED_LIMIT 作为启发式函数。