我在SO看过其他有关此事的帖子。
我试图找到图表中节点之间的最短路径。节点之间的每条边都有一个(X, Y)坐标。
我想计算从Node I到Node J的最短距离。一旦有了，我想从最短路径的坐标中加上所有的X值和Y值。
我已经被困在这里好几个小时，并且会喜欢一些见解。
以下是代码：
class Vertex implements Comparable<Vertex> {
private int id;
private List<Edge> adjacencyList;
private Vertex previousVertex;
private double minDistance;
private Coordinate point;
public Vertex(int id, Coordinate point) {
this.id = id;
this.point = point;
this.adjacencyList = new ArrayList<>();
}
public int getID() {
return this.id;
}
public Coordinate getPoint() {
return this.point;
}
public List<Edge> getAdjacencyList() {
return this.adjacencyList;
}
public void addNeighbour(Edge edge) {
this.adjacencyList.add(edge);
}
public Vertex getPreviousVertex() {
return this.previousVertex;
}
public void setPreviousVertex(Vertex previousVertex) {
this.previousVertex = previousVertex;
}
public double getMinDistance() {
return this.minDistance;
}
public void setMinDistance(double minDistance) {
this.minDistance = minDistance;
}
public int compareTo(Vertex other) {
return Double.compare(this.minDistance, other.minDistance);
}
}
class Edge {
private double weight;
private Vertex targetVertex;
public Edge(double weight, Vertex targetVertex) {
this.weight = weight;
this.targetVertex = targetVertex;
}
public double getWeight() {
return this.weight;
}
public void setWeight(double weight) {
this.weight = weight;
}
public Vertex getTargetVertex() {
return this.targetVertex;
}
public void setTargetVertex(Vertex targetVertex) {
this.targetVertex = targetVertex;
}
}
class Algorithm {
public void shortestPath(Vertex startVertex) {
startVertex.setMinDistance(0);
PriorityQueue<Vertex> queue = new PriorityQueue<>();
queue.add(startVertex);
while (!queue.isEmpty()) {
Vertex actualVertex = queue.poll();
for (Edge edge : actualVertex.getAdjacencyList()) {
Vertex v = edge.getTargetVertex();
double weight = edge.getWeight();
double currentDistance = actualVertex.getMinDistance() + weight;
if (currentDistance < v.getMinDistance()) {
queue.remove(v);
v.setMinDistance(currentDistance);
v.setPreviousVertex(actualVertex);
queue.add(v);
}
}
}
}
public List<Vertex> getShortestPathTo(Vertex targetVertex){
List<Vertex> path = new ArrayList<Vertex>();
for (Vertex vertex = targetVertex; vertex != null; vertex = vertex.getPreviousVertex()){
path.add(vertex);
}
Collections.reverse(path);
return path;
}
}
class Coordinate {
private int x;
private int y;
Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
public int getX() {
return this.x;
}
public int getY() {
return this.y;
}
public static Coordinate readInput(Scanner in) {
String[] temp = in.nextLine().split(" ");
return new Coordinate(Integer.parseInt(temp[0]), Integer.parseInt(temp[1]));
}
}
如果我从这个文本文件中读取
3 //坐标数
0 0 //（x，y）
1 1 //（x，y）
2 0 //（x，y）
1 2 //坐标1到2之间的边缘
2 3 //坐标2到3之间的边缘
我的测试用例如下：
class Test {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String[] constants = s.nextLine().split(" ");
final int N = Integer.parseInt(constants[0]);
List<Coordinate> junctions = new ArrayList<>();
List<Coordinate> paths = new ArrayList<>();
List<Vertex> vertices = new ArrayList<>();
for(int i = 0; i < N; i++) {
junctions.add(Coordinate.readInput(s));
}
for(int i = 0; i < N-1; i++) {
paths.add(Coordinate.readInput(s));
}
for(int i = 0; i < N-1; i++) {
int x = junctions.get(paths.get(i).getX() - 1).getX();
int x1 = junctions.get(paths.get(i).getY() - 1).getX();
int y = junctions.get(paths.get(i).getX() - 1).getY();
int y1 = junctions.get(paths.get(i).getY() - 1).getY();
Vertex vertex1 = new Vertex(paths.get(i).getX(), new Coordinate(x, y));
Vertex vertex2 = new Vertex(paths.get(i).getY(), new Coordinate(x1, y1));
double distance = Math.sqrt(Math.pow(x - x1, 2) + Math.pow(y - y1, 2));
vertex1.addNeighbour(new Edge(distance, vertex2));
vertices.add(vertex1);
vertices.add(vertex2);
}
Algorithm a = new Algorithm();
int x = 0;
int y = 0;
for(int i = 0; i < vertices.size(); i++) {
a.shortestPath(vertices.get(i));
for(Vertex vertex : a.getShortestPathTo(vertices.get(i+1))) {
x += vertices.get(vertex.getID()).getPoint().getX();
y += vertices.get(vertex.getID()).getPoint().getY();
}
}
//This prints out "Total X: 5 Total Y: 3" (should be 3 and 1)
System.out.println("Total X: " + x + " Total Y: " + y);
}
}
答案 0 :(得分：1)
要解决此问题，您需要跟踪每个节点，Dijktra树中该节点的父节点是什么。一旦你跟踪了你能够以递归方式恢复最短路径，就可以遍历它并计算你需要知道的东西。
答案 1 :(得分：1)
你的问题在于这一部分：
public void shortestPath(Vertex startVertex) {
startVertex.setMinDistance(0);
PriorityQueue<Vertex> queue = new PriorityQueue<>();
queue.add(startVertex);
//The rest is omitted
}
每次运行shortestPath方法时，都应将所有顶点中的所有minDistance重置为无穷大，而不仅仅是startVertex。
对于除startVertex之外的所有顶点，开头的minDistance应设置为无穷大（或Double.MAX_VALUE），或者始终为0。
代码：
for(Vertex v : vertices){
v.setMinDistance(Double.MAX_VALUE);
v.setPreviousVertex(null);
}
a.shortestPath(vertices.get(i));
此外，在类Test的第三个循环中，您初始化多个相同的顶点。所以，你应该做的是，预先初始化所有顶点并将它们保存在这样的数组中：
for(int i = 0; i < N; i++){
vertices.add(new Vertex(i + 1, junctions.get(i));
}
for(int i = 0; i < N - 1; i++){
//Initialize x, y, x1, y1 here, I just skipped it
Vertex vertex1 = vertices.get(paths.getX() - 1);
Vertex vertex2 = vertices.get(paths.getY() - 1);
double distance = Math.sqrt(Math.pow(x - x1, 2) + Math.pow(y - y1, 2));
vertex1.addNeighbour(new Edge(distance, vertex2));
}
答案 2 :(得分：1)
正如所承诺的那样，我为我的二分图解算器创建了一个github repo。你可以看看这里：Bipartite Solver on GitHub
正如您所解释的那样，您有一组点，您正试图找到它们之间的最小距离。你有一大堆X和Y坐标，并且你试图找到节点之间最小的欧氏距离。
你最感兴趣的算法是Jonker-Volgenant，它实现了Dijkstra的遍历节点的算法。
该库包含一堆standalone executables，您可以运行它来测试代码。
请告诉我这是否对您有用。