A* 搜索中的“跳跃”

Question

我正在尝试实现 A* 搜索算法，这是我的尝试：

void Map::findPath(Robot robot, Model target, vector<Node> nodeList, vector<Node> &closedList) {
Node targetNode = target.getCurrentNode();
Node startNode = robot.getCurrentNode();

vector<Node> openList;

openList.push_back(startNode);
startNode.setG(0);startNode.setH(0);startNode.setF(0);
int i = 0;
while (!openList.empty()) {
    Node currentNode = nodeWithLowestFScore(openList);//always returns the most recent one
    /*for ( int i = 0; i < openList.size(); i++){
        cout << "X: " << openList[i].getX() << " Y: " << openList[i].getY() <<
                " G: " << openList[i].getG() << " M: " << openList[i].getH() << endl;
    }*/

    cout << i++ << ". " << currentNode.getX() << " " << currentNode.getY() << " G: " <<
            currentNode.getG() << " M: " << currentNode.getH() << endl;

    closedList.push_back(currentNode);
    removeFromVector(openList, currentNode);
    if (inVector(closedList, targetNode))
        break;
    vector<Node> adjacentNodes;
    currentNode.getWalkableAdjacentNodes(nodeList, adjacentNodes);
    for ( int i = 0; i < adjacentNodes.size(); i++){
        if (inVector(closedList, adjacentNodes[i]))
            continue;
        if (!inVector(openList, adjacentNodes[i])){
            adjacentNodes[i].setParent(&currentNode);
            adjacentNodes[i].setG(currentNode.getG() +1);//distance is always 1 between adjacent nodes
            adjacentNodes[i].setH(adjacentNodes[i].getDistance(targetNode, 'm'));//heuristic as manhattan
            adjacentNodes[i].setF(adjacentNodes[i].getG() + adjacentNodes[i].getH());
            openList.push_back(adjacentNodes[i]);
        }
        if (inVector(openList, adjacentNodes[i])){//update if it's in the list already
            //?

        }
    }
}

}

我认为函数的名称是不言自明的，所以我不会深入探讨它们。无论如何，在我的示例输出中，我试图从 (x:0, y:-2) 转到 (x:-7, y:6)

1. 0 -2
2. -1 -2
3. -2 -2
4. -3 -2
5. -3 -1
6. -3 0
7. -4 0
8. -5 0
9. -5 1
10. -5 2
11. -5 3
12. -5 4
13. -6 4
14. -7 4
15. -5 5
16. -5 6
17. -3 1
18. -3 2
19. -3 3
20. -3 4
21. -2 4
22. -2 2
23. -4 6
24. -5 7
25. -8 4
26. -4 4
27. -5 -1
28. -1 -3
29. 1 -2
30. 2 -2
31. -1 -4
32. -2 -4
33. -3 -4
34. -5 -2
35. -9 4
36. -9 5
37. -9 6
38. -8 6
39. -7 6

在第 14 行之前一切似乎都很顺利，但随后突然跳转到 (5,5)。 非常感谢任何帮助。

Answer 1

访问节点的顺序不一定是您的最短路径。据我所知，该算法运行良好。观察到节点 -5 4 在第 12 行被访问，因此 -5 5 只是在第 15 行访问的那个节点的邻居。要获得最短路径，您应该将最终节点的 parentNode 追溯到初始算法结束时的节点。