我一直在努力让这个算法运行起来，但我不能为我的生活做准备。我在网上阅读了很多教程，并在AS3，javascript和C ++中阅读了源代码;但我无法适应我所看到的自己的代码。
我创建了一个AStar类，它有一个名为Node的嵌套类。地图是名为MAP的2D数组。
我遇到的最大问题是在pathfind函数中拉出F值。
我已经实现了F = G + H，我的问题是实际的AStar算法。有人可以请求帮助，这是我到目前为止还有多远：
import java.util.ArrayList;
public class AStar
{
int MAP[][];
Node startNode, endNode;
public AStar(int MAP[][], int startXNode, int startYNode,
int endXNode, int endYNode)
{
this.MAP = MAP;
startNode = new Node(startXNode, startYNode);
endNode = new Node(endXNode, endYNode);
}
public void pathfinder()
{
ArrayList openList = new ArrayList();
ArrayList closedList = new ArrayList();
}
public int F(Node startNode, Node endNode)
{
return (H(startNode, endNode) + G(startNode));
}
//H or Heuristic part of A* algorithm
public int H(Node startNode, Node endNode)
{
int WEIGHT = 10;
int distance = (Math.abs(startNode.getX() - endNode.getX()) + Math.abs(startNode.getY() - endNode.getY()));
return (distance * WEIGHT);
}
public int G(Node startNode)
{
if(MAP[startNode.getX() - 1][startNode.getY()] != 1)
{
return 10;
}
if(MAP[startNode.getX() + 1][startNode.getY()] != 1)
{
return 10;
}
if(MAP[startNode.getX()][startNode.getY() -1] != 1)
{
return 10;
}
if(MAP[startNode.getX()][startNode.getY() + 1] != 1)
{
return 0;
}
return 0;
}
public class Node
{
private int NodeX;
private int NodeY;
private int gScore;
private int hScore;
private int fScore;
public Node(int NodeX, int NodeY)
{
this.NodeX = NodeX;
this.NodeY = NodeY;
}
public int getX()
{
return NodeX;
}
public int getY()
{
return NodeY;
}
public int getG()
{
return gScore;
}
public void setG(int gScore)
{
this.gScore = gScore;
}
public int getH()
{
return hScore;
}
public void setH(int hScore)
{
this.hScore = hScore;
}
public int getF()
{
return fScore;
}
public void setF(int fScore)
{
this.fScore = fScore;
}
}
}
这是我使用探路者功能所能达到的最远：
public void pathfinder()
{
LinkedList<Node> openList = new LinkedList();
LinkedList<Node> closedList = new LinkedList();
Node currentNode;
openList.add(startNode);
while(openList.size() > 0)
{
currentNode = (Node) openList.get(0);
closedList.add(currentNode);
for(int i = 0; i < openList.size(); i++)
{
int cost = F(currentNode, endNode);
}
}
}
答案 0 :(得分：6)
我最近将这个A *代码放在一起以解决Project Euler问题。您必须填写Node个对象矩阵的详细信息。使用它需要您自担风险，但我可以说它解决了问题：）
public class Node {
List<Node> neighbors = new ArrayList<Node>();
Node parent;
int f;
int g;
int h;
int x;
int y;
int cost;
}
public List<Node> aStar(Node start, Node goal) {
Set<Node> open = new HashSet<Node>();
Set<Node> closed = new HashSet<Node>();
start.g = 0;
start.h = estimateDistance(start, goal);
start.f = start.h;
open.add(start);
while (true) {
Node current = null;
if (open.size() == 0) {
throw new RuntimeException("no route");
}
for (Node node : open) {
if (current == null || node.f < current.f) {
current = node;
}
}
if (current == goal) {
break;
}
open.remove(current);
closed.add(current);
for (Node neighbor : current.neighbors) {
if (neighbor == null) {
continue;
}
int nextG = current.g + neighbor.cost;
if (nextG < neighbor.g) {
open.remove(neighbor);
closed.remove(neighbor);
}
if (!open.contains(neighbor) && !closed.contains(neighbor)) {
neighbor.g = nextG;
neighbor.h = estimateDistance(neighbor, goal);
neighbor.f = neighbor.g + neighbor.h;
neighbor.parent = current;
open.add(neighbor);
}
}
}
List<Node> nodes = new ArrayList<Node>();
Node current = goal;
while (current.parent != null) {
nodes.add(current);
current = current.parent;
}
nodes.add(start);
return nodes;
}
public int estimateDistance(Node node1, Node node2) {
return Math.abs(node1.x - node2.x) + Math.abs(node1.y - node2.y);
}
答案 1 :(得分：0)
我不知道你是否只想使用简单的类型，或者如果你只是没有考虑它，但是你需要有一个PriorityQueue来让你的A *工作。
一种好的思考方式是将起始点放入距离为0的优先级队列中，然后启动一个仅在先前队列为空时停止的循环。
在循环中，你取出最小节点，并检查它是否之前没有打开，或者如果它已经打开，如果你现在找到了一个更短的方法。 如果其中任何一个为真，则将距离添加到新节点，将边/边距添加到地图，然后将距离+启发式添加到优先级队列。
我已经写过这个用于布尔网格，以及1D和2D数组之间的持续转换，但我希望它是可读的：
public void AStarRoute()
{
gridDist = new double[rows][cols];
System.out.println("Start of AStarRoute");
MinPriorityQueue pq = new MinPriorityQueue(rows * cols);
edgeTo = new HashMap<Integer, Integer>();
gridDist[x1Dto2D(start)][y1Dto2D(start)] = 0;
pq.insert(start, 0);
int from;
while (!pq.isEmpty()) {
from = pq.delMin();
int x = x1Dto2D(from);
int y = y1Dto2D(from);
for (int i = -1; i <= 1; i++) {
for (int j = -1; j <= 1; j++) {
int newX = x + i;
int newY = y + j;
if (newX >= 0 && newY >= 0 && newX < cols && newY < rows && !(i == 0 && j == 0)) {
if (grid[newX][newY]) {
//System.out.println("NewDist: " + gridDist[newX][newY] + " - OldDist+dist: " + (gridDist[x][y] + ((Math.abs(i) == Math.abs(j)) ? 1.4 : 1.0)) + ":" + (int)(gridDist[x][y] + ((Math.abs(i) == Math.abs(j)) ? 1.4 : 1.0)));
if (!edgeTo.containsKey(convert2Dto1D(newX, newY)) || gridDist[newX][newY] > (gridDist[x][y] + ((Math.abs(i) == Math.abs(j)) ? 14 : 10))) {
gridDist[newX][newY] = (int)(gridDist[x][y] + ((Math.abs(i) == Math.abs(j)) ? 14 : 10));
maxDistToEnd = (int)Math.max(maxDistToEnd, gridDist[newX][newY]);
edgeTo.put(convert2Dto1D(newX, newY), convert2Dto1D(x, y));
pq.insert(convert2Dto1D(newX, newY), gridDist[newX][newY] + (int)Math.sqrt(Math.pow((newX - x1Dto2D(end))*10, 2) + Math.pow((newY - y1Dto2D(end))*10, 2)));
if(convert2Dto1D(newX, newY) == end){
System.out.println("End found at (" + newX + ", " + newY + ")");
paintGridDist = true;
route = new ArrayList<Integer>();
int n = convert2Dto1D(newX, newY);
route.add(n);
do{
n = edgeTo.get(n);
route.add(n);
}while(start != n);
repaint();
return;
}
}
}
}
}
}
}
paintGridDist = true;
repaint();
}