【发布时间】：2016-10-23 06:58:34
【问题描述】：
编辑：
为了帮助这个问题的观众成为回答者，我会注意到问题似乎出在 Jump 的对角线情况下。该方法中的什么在递归运行时会导致速度变慢？
编辑结束。
在我正在开发的 XNA 游戏中，我使用 A* + JPS 算法在每帧的统一方形网格上导航。我了解了 JPS here 和 here。但是，当我运行游戏时，帧速率会下降。帧率太低了，以至于无法玩游戏。删除对 Jump Point Search 'jump' 的调用，并改为使用常规 A*，解决了足够大的正方形大小的问题。根据这篇文章，JPS 应该比普通的 A* 效率高得多。速度变慢的原因似乎是“Diagonal Case”中对 Jump 的两次调用（Jump 方法的第 15-29 行）。
显然，我的实现中一定有一些错误/效率低下。但它是什么？
代码：
Jump方法是JPS递归跳转函数的实现。
public Vector2? Jump(Vector2 current, Vector2 direction, Vector2 start, Vector2 end)
{
// Position of new node we are going to consider:
Vector2 next = current + direction * SquareSize;
// If it's blocked we can't jump here
if (IsBlocked(next))
return null;
// If the node is the goal return it
if (next.X == end.X && next.Y == end.Y)
return next;
// Diagonal Case
if (direction.X != 0 && direction.Y != 0)
{
Vector2 horizontalBehind = current - new Vector2(direction.X * SquareSize, 0);
Vector2 verticalBehind = current - new Vector2(0, direction.Y * SquareSize);
if (IsBlocked(horizontalBehind) || IsBlocked(verticalBehind))
{
return current;
}
// Check in horizontal and vertical directions for forced neighbors
// This is a special case for diagonal direction
if (Jump(next, new Vector2(direction.X, 0), start, end) != null || Jump(next, new Vector2(0, direction.Y), start, end) != null)
{
return current;
}
}
else
{
// Horizontal case
if (direction.X != 0)
{
Vector2 topBehind = current - new Vector2(direction.X * SquareSize, SquareSize);
Vector2 above = current - new Vector2(0, SquareSize);
Vector2 bottomBehind = current + new Vector2(-direction.X * SquareSize, SquareSize);
Vector2 below = current + new Vector2(0, SquareSize);
if (IsBlocked(topBehind) || IsBlocked(above) || IsBlocked(bottomBehind) || IsBlocked(below))
{
return current;
}
}
else
{
Vector2 leftBehind = current - new Vector2(SquareSize, direction.Y * SquareSize);
Vector2 left = current - new Vector2(SquareSize, 0);
Vector2 rightBehind = current + new Vector2(-SquareSize, direction.Y * SquareSize);
Vector2 right = current - new Vector2(SquareSize, 0);
if (IsBlocked(leftBehind) || IsBlocked(left) || IsBlocked(rightBehind) || IsBlocked(right))
{
return current;
}
}
}
// If forced neighbor was not found try next jump point
return Jump(next, direction, start, end);
}
#endregion
}
Navigate是实现A*的方法，从'start'参数到'end'参数。
PriorityQueue 是基于this article 的通用实现。它的 Enqueue 和 Dequeue 方法具有 O(log2(n)) 复杂度。
public Vector2? Navigate(Vector2 start, Vector2 end)
{
PriorityQueue<float, Vector2> openSet = new PriorityQueue<float, Vector2>();
List<Vector2> closedSet = new List<Vector2>(10);
Dictionary<Vector2, Vector2> cameFrom = new Dictionary<Vector2, Vector2>(10);
Dictionary<Vector2, float> gScores = new Dictionary<Vector2, float>(10);
gScores[start] = 0;
openSet.Enqueue(H(start, end), start);
while (openSet.Count != 0)
{
Vector2 current = openSet.Dequeue().Value;
if (WorldMap.InSquare(current) == WorldMap.InSquare(end))
return ReconstructPath(cameFrom, current, start);
List<Vector2> neighbours = WorldMap.GetNeighbours(current, start, end);
closedSet.Add(current);
foreach(Vector2 neighbour in neighbours)
{
if (closedSet.Contains(neighbour))
continue;
float tenativeGScore = gScores[current] + Vector2.Distance(current, neighbour);
if(!gScores.ContainsKey(neighbour) || gScores[neighbour] > tenativeGScore)//Discover a new node || Find a better path to a node
{
cameFrom[neighbour] = current;
gScores[neighbour] = tenativeGScore;
float fScore = tenativeGScore + H(neighbour, end);//Calculate F.
openSet.Enqueue(fScore, neighbour);
}
}
}
return null;
}
GetNeighbours 是一种返回节点“向量”的邻居的方法。 A* 版本：
public List<Vector2> GetNeighbours(Vector2 point, Vector2 start, Vector2 end)
{
Vector2[] directions = new Vector2[8];
List<Vector2> neighbours = new List<Vector2>(8);
directions[0] = Vector2.UnitX;//right
directions[1] = -Vector2.UnitX;//left
directions[2] = Vector2.UnitY;//down
directions[3] = -Vector2.UnitY;//up
directions[4] = Vector2.UnitX + Vector2.UnitY;//down right
directions[5] = -Vector2.UnitX + Vector2.UnitY;//down left
directions[6] = Vector2.UnitX - Vector2.UnitY;//up right
directions[7] = -Vector2.UnitX - Vector2.UnitY;//up left
foreach(Vector2 direction in directions)
{
Vector2 neighbour = point + direction * SquareSize;
if (!IsBlocked(neighbour))
neighbours.Add(neighbour);
}
return neighbours;
}
跳转点搜索版本：
public List<Vector2> GetNeighbours(Vector2 point, Vector2 start, Vector2 end)
{
Vector2[] directions = new Vector2[8];
List<Vector2> neighbours = new List<Vector2>(8);
directions[0] = Vector2.UnitX;//right
directions[1] = -Vector2.UnitX;//left
directions[2] = Vector2.UnitY;//down
directions[3] = -Vector2.UnitY;//up
directions[4] = Vector2.UnitX + Vector2.UnitY;//down right
directions[5] = -Vector2.UnitX + Vector2.UnitY;//down left
directions[6] = Vector2.UnitX - Vector2.UnitY;//up right
directions[7] = -Vector2.UnitX - Vector2.UnitY;//up left
foreach(Vector2 direction in directions)
{
//The only difference between this GetNeighbours and the other one
//is that this one calls Jump here.
Vector2? jp = Jump(point + direction * SquareSize, direction, start, end);
if (jp != null)
neighbours.Add((Vector2)jp);
}
return neighbours;
}
InSqaure 是一种返回 Vector2 的方法，它表示 Vector2 所在的正方形。它的复杂度为 O(1)。
IsBlocked 是一种检查 Vector2 是否在地图内的方法，并且它也在一个被阻挡的方块中（“阻挡”意味着一个方块中有障碍物）。它的复杂度为 O(log2(n))。
其他信息：
- 我目前只在目标和起点之间没有障碍物的情况下检查这个。这意味着 JPS 算法仅扩展 1 个正方形。同时，正则 A* 扩展 8 以找到从 (100, 100) 到 (0, 0) 的路径，但在找到 (100, 100) 到 (-800, -580) 之间的距离时是 2325 个正方形。在最后一种情况下，会出现明显的帧速率下降，导致游戏几乎无法玩。
- 帧速率已解锁。
- 在小于约 1200 个方格的网格中没有可见的卡顿，游戏有轻微的可接受的帧率下降，最高可达 1600 左右。
如果需要更多信息，我会很乐意提供，
提前致谢！
【问题讨论】：
标签： c# algorithm performance xna a-star