计算矩阵中非直接连接节点之间的距离 [英] Calculating distance between non directly-connected nodes in matrix
本文介绍了计算矩阵中非直接连接节点之间的距离的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我为城市建立了邻接矩阵，并在它们之间建立了联系.例如A-B，B-C，C-D.现在，我想知道是否可以计算未连接的城市之间的距离.是否可以计算未连接节点之间的矩阵距离并找到路径?
I made a adjacency matrix for cities and connecting between them. And for example A-B, B-C, C-D. Now I am wonder if I can calculate distance between cities that aren't connected. Is it possible to calculate distance in matrix between non connected nodes and find path?
城市班
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
public class GraphCities {
private List<String> cities;
private int[][] matrix;
Scanner s = new Scanner(System.in);
public GraphCities() {
this.cities = new LinkedList<String>();
}
public void addCity(String name) {
if (!cities.contains(name)) {
cities.add(name);
} else {
System.out.println("City " + name + " is already added.");
}
}
public void makeGraph() {
System.out
.println("Distance between cities, if they aren't connected insert -1");
matrix = new int[cities.size()][cities.size()];
for (int i = 0; i < matrix.length; i++) {
for (int j = i; j < matrix.length; j++) {
if (i == j) {
matrix[i][j] = 0;
} else {
System.out.println("Distance from "
+ cities.get(i) + " to " + cities.get(j));
int distance = s.nextInt();
matrix[i][j] = distance;
matrix[j][i] = distance;
}
}
}
}
public void show() {
String show = "\t";
for (int i = 0; i < cities.size(); i++) {
show += cities.get(i) + "\t";
}
show += "\n";
for (int i = 0; i < matrix.length; i++) {
show += cities.get(i) + "\t";
for (int j = 0; j < matrix.length; j++) {
if (matrix[i][j] != -1) {
show += matrix[i][j] + "\t";
} else {
show += "-\t";
}
}
show += "\n";
}
System.out.println(show);
}
}
主要方法
public class Main {
public static void main(String[] args) {
GraphCities c = new GraphCities();
c.addCity("A");
c.addCity("B");
c.addCity("C");
c.addCity("D");
c.addCity("E");
c.makeGraph();
System.out.println();
c.show();
}
}
这是我运行main方法时的输出，我认为一切正常.
This is my output when i run main method and i think everything is ok.
A B C D E
A 0 50 - - -
B 50 0 30 - -
C - 30 0 40 -
D - - 40 0 20
E - - - 20 0
现在我想计算从A到D的距离，但是我不知道该怎么做.我将不胜感激.谢谢！
Now I want to calculate distance from A to D, but i haven't any idea how to do it. I will appreciate any help. Thanks!
推荐答案
要在加权图中查找最短路径(路线的每个部分的权重都不同)，可以使用 mcve .可以将其复制粘贴到一个文件(Main.java)中并执行.
(此答案基于编辑问题之前发布的代码)
请注意评论:
To find the shortest path in a weighted graph (each part of the route has a different weight) you can use Dijkstra's Shortest Path Algorithm.
The following code is a one-file mcve. It can be copy-pasted into one file (Main.java) , and executed.
(This answer is base on the code posted before editing the question)
Please note the comments:
import java.util.ArrayList;
import java.util.Collection;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
public class Main {
public static void main(String[] args) {
AdjList aList = new AdjList();
CityNode a = new CityNode("A");
CityNode b = new CityNode("B");
CityNode c = new CityNode("C");
CityNode d = new CityNode("D");
CityNode e = new CityNode("E");
aList.addCity(a);
aList.addCity(b);
aList.addCity(c);
aList.addCity(d);
aList.addCity(e);
aList.connectCities(a, b, 50);
aList.connectCities(b, c, 30);
aList.connectCities(c, d, 40);
aList.connectCities(d, e, 20);
aList.show();
FindPath findPath = new FindPath();
System.out.println(findPath.calculateShortestPath(aList, a, e)); //prints 140 as expected
//add some complexity
CityNode f = new CityNode("F");
aList.addCity(f);
aList.connectCities(b, f, 10);
aList.connectCities(f, d, 10);
System.out.println(findPath.calculateShortestPath(aList, a, e));//prints 90 as expected
}
}
class FindPath{
//map to hold distances of all node from origin. at the end this map should contain
//the shortest distance between origin (from) to all other nodes
Map<CityNode, Integer> distances;
//using Dijkstra algorithm
public int calculateShortestPath(AdjList aList, CityNode from, CityNode to) {
//a container to hold which cities the algorithm has visited
Set<CityNode> settledCities = new HashSet<>();
//a container to hold which cities the algorithm has to visit
Set<CityNode> unsettledCities = new HashSet<>();
unsettledCities.add(from);
//map to hold distances of all node from origin. at the end this map should contain
//the shortest distance between origin (from) to all other nodes
distances = new HashMap<>();
//initialize map with values: 0 distance to origin, infinite distance to all others
//infinite means no connection between nodes
for(CityNode city :aList.getCities()){
int distance = city.equals(from) ? 0 : Integer.MAX_VALUE;
distances.put(city, distance);
}
while (unsettledCities.size() != 0) {
//get the unvisited city with the lowest distance
CityNode currentCity = getLowestDistanceCity(unsettledCities);
//remove from unvisited, add to visited
unsettledCities.remove(currentCity); settledCities.add(currentCity);
Map<CityNode, Integer> connectedCities = currentCity.getConnectedCities();
for( CityNode city : connectedCities.keySet()){
//check if new distance is shorted than the previously found distance
if(distances.get(currentCity) + connectedCities.get(city) < distances.get(city)){
//if so, keep the shortest distance found
distances.put(city, distances.get(currentCity) + connectedCities.get(city));
//if city has not been visited yet, add it to unsettledCities
if(! settledCities.contains(city)) {
unsettledCities.add(city);
}
}
}
}
return distances.get(to);
}
private CityNode getLowestDistanceCity(Set <CityNode> unsettledCities) {
return unsettledCities.stream()
.min((c1,c2)-> Integer.compare(distances.get(c1), distances.get(c2)))
.orElse(null);
}
}
class CityNode {
private static int counter =0;
private final String name;
//assign unique id to each node. safer than to rely on unique name
private final int id = counter ++;
//map to hold all connected cities and distance to each
private final Map<CityNode, Integer> connectedCities;
public CityNode(String name) {
super();
this.name = name;
connectedCities = new HashMap<>();
}
public String getName() {
return name;
}
//not null safe. distance must not be negative
public void connect(CityNode connectTo, int distance) {
if(connectTo.equals(this)) throw new IllegalArgumentException("Node can not connect to istself");
connectedCities.put(connectTo, distance);
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder(name);
sb.append(connectedCities.keySet().isEmpty() ? " not connected" : " conntected to: " );
for ( CityNode city : connectedCities.keySet()) {
sb.append(city.getName()).append("-")
.append(connectedCities.get(city)).append("km ");
}
return sb.toString();
}
public int getId() {
return id;
}
public Map<CityNode, Integer> getConnectedCities(){
return connectedCities;
}
@Override
public boolean equals(Object o) {
if(o == null ||!(o instanceof CityNode)) return false;
CityNode c = (CityNode) o;
return c.getName().equalsIgnoreCase(name) && id == c.getId();
}
@Override
public int hashCode() {
int hash = 31 * 7 + id;
return name == null ? hash : name.hashCode();
}
}
class AdjList {
//use set which prevents duplicate entries
private final Set<CityNode> citiesList;
public AdjList() {
citiesList = new HashSet<>();
}
//adds city if is not already present. returns true if city was added
public boolean addCity(CityNode city) {
return citiesList.add(city);
}
//not null safe
public void connectCities(CityNode city1, CityNode city2, int distance) {
//assuming undirected graph
city1.connect(city2, distance);
city2.connect(city1, distance);
}
public CityNode getCityByName(String name) {
for (CityNode city : citiesList) {
if (city.getName().equalsIgnoreCase(name))
return city;
}
return null;
}
public void show() {
for (CityNode city : citiesList) {
System.out.println(city);
}
}
//get a copy of cities list
public Collection<CityNode> getCities(){
return new ArrayList<>(citiesList);
}
}
如果您需要不使用Stream的getLowestDistanceCity版本，请使用:
If you need a version of getLowestDistanceCity which does not use Stream use :
private CityNode getLowestDistanceCity(Set <CityNode> unsettledCities) {
CityNode lowestDistanceCity = null;
int lowestDistance = Integer.MAX_VALUE;
for (CityNode city: unsettledCities) {
int nodeDistance = distances.get(city);
if (nodeDistance < lowestDistance) {
lowestDistance = nodeDistance;
lowestDistanceCity = city;
}
}
return lowestDistanceCity;
}
编辑:使用邻接矩阵的实现看起来像这样:
an implementation using adjacency matrix could look like this:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
public class DijkstraAdjacencyMatrix {
public static void main(String[] args) {
Set<CityNode> cities = new HashSet<>();
CityNode a = new CityNode("A");
CityNode b = new CityNode("B");
CityNode c = new CityNode("C");
CityNode d = new CityNode("D");
CityNode e = new CityNode("E");
cities.addAll(List.of(a,b,c,d,e));
CitiesGraph graph = new CitiesGraph(cities);
graph.connectCities(a, b, 50);
graph.connectCities(b, c, 30);
graph.connectCities(c, d, 40);
graph.connectCities(d, e, 20);
graph.show();
FindPath findPath = new FindPath();
System.out.println(findPath.calculateShortestPath(graph, a, e)); //prints 140 as expected
//to add some complexity we need to construct a new CitiesGraph. It is not reusable
CityNode f = new CityNode("F");
cities.add(f);
graph = new CitiesGraph(cities);
graph.connectCities(a, b, 50);
graph.connectCities(b, c, 30);
graph.connectCities(c, d, 40);
graph.connectCities(d, e, 20);
graph.connectCities(b, f, 10);
graph.connectCities(f, d, 10);
graph.show();
System.out.println(findPath.calculateShortestPath(graph, a, e));//prints 90 as expected
}
}
class FindPath{
//map to hold distances of all node from origin. at the end this map should contain
//the shortest distance between origin (from) to all other nodes
Map<CityNode, Integer> distances;
//using Dijkstra algorithm
public int calculateShortestPath(CitiesGraph graph, CityNode from, CityNode to) {
//a container to hold which cities the algorithm has visited
Set<CityNode> settledCities = new HashSet<>();
//a container to hold which cities the algorithm has to visit
Set<CityNode> unsettledCities = new HashSet<>();
unsettledCities.add(from);
//map to hold distances of all node from origin. at the end this map should contain
//the shortest distance between origin (from) to all other nodes
distances = new HashMap<>();
//initialize map with values: 0 distance to origin, infinite distance to all others
//infinite means no connection between nodes
for(CityNode city :graph.getCities()){
int distance = city.equals(from) ? 0 : Integer.MAX_VALUE;
distances.put(city, distance);
}
while (unsettledCities.size() != 0) {
//get the unvisited city with the lowest distance
CityNode currentCity = getLowestDistanceCity(unsettledCities);
//remove from unvisited, add to visited
unsettledCities.remove(currentCity); settledCities.add(currentCity);
Collection<CityNode> connectedCities = graph.getCitiesConnectedTo(currentCity);
//iterate over connected city to update distance to each
for( CityNode city : connectedCities){
//check if new distance is shorted than the previously found distance
int distanceToCity = graph.getDistanceBetween(city, currentCity);
if(distanceToCity <= 0) {
continue;
}
if(distances.get(currentCity) + distanceToCity < distances.get(city)){
//if so, keep the shortest distance found
distances.put(city,distances.get(currentCity) + distanceToCity);
//if city has not been visited yet, add it to unsettledCities
if(! settledCities.contains(city)) {
unsettledCities.add(city);
}
}
}
}
return distances.get(to);
}
private CityNode getLowestDistanceCity(Set <CityNode> unsettledCities) {
return unsettledCities.stream()
.min((c1,c2)-> Integer.compare(distances.get(c1), distances.get(c2)))
.orElse(null);
}
}
class CityNode {
private static int counter =0;
private final String name;
//assign unique id to each node. safer than to rely on unique name
private final int id = counter ++;
public CityNode(String name) {
this.name = name;
}
public String getName() {
return name;
}
@Override
public String toString() {
StringBuilder sb = new StringBuilder("City ");
sb.append(name).append(" (id=").append(id).append(")");
return sb.toString();
}
public int getId() {
return id;
}
@Override
public boolean equals(Object o) {
if(o == null || !(o instanceof CityNode)) return false;
CityNode c = (CityNode) o;
return c.getName().equalsIgnoreCase(name) && id == c.getId();
}
@Override
public int hashCode() {
int hash = 31 * 7 + id;
return name == null ? hash : name.hashCode();
}
}
class CitiesGraph{
//use set which prevents duplicate entries
private final Set<CityNode> cities;
private final int[][] adjacencyMatrix;
private static final int NOT_CONNECTED = -1;
public CitiesGraph(Set<CityNode> cities) {
this.cities = cities;
adjacencyMatrix = new int[cities.size()][cities.size()];
//initialize matrix
for(int row = 0; row < adjacencyMatrix.length ; row++){
for(int col = 0; col < adjacencyMatrix[row].length ; col++){
adjacencyMatrix[row][col] = row == col ? 0 : NOT_CONNECTED ;
}
}
}
public void connectCities(CityNode city1, CityNode city2, int distance) {
//assuming undirected graph
adjacencyMatrix[city1.getId()][city2.getId()] = distance;
adjacencyMatrix[city2.getId()][city1.getId()] = distance;
}
public int getDistanceBetween(CityNode city1, CityNode city2) {
return adjacencyMatrix[city1.getId()][city2.getId()];
}
public Collection<CityNode> getCitiesConnectedTo(CityNode city) {
Collection<CityNode> connectedCities = new ArrayList<>();
//iterate over row representing city's connections
int column = 0;
for(int distance : adjacencyMatrix[city.getId()]){
if(distance != NOT_CONNECTED && distance > 0) {
connectedCities.add(getCityById(column));
}
column++;
}
return connectedCities;
}
public CityNode getCityById(int id) {
for (CityNode city : cities) {
if (city.getId() == id) return city;
}
return null;
}
public void show() {
for(int[] row : adjacencyMatrix){
System.out.println(Arrays.toString(row));
}
}
//get a copy of cities list
public Collection<CityNode> getCities(){
return new ArrayList<>(cities);
}
}
这篇关于计算矩阵中非直接连接节点之间的距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
