1
Je résous le juge en ligne de sphère shortest path problem. Ce morceau de code me donne du mal:résoudre cette NullPointerException sur binarySearch de java
int sourceIndex = Arrays.binarySearch(citiesIds,source);
int destinationIndex= Arrays.binarySearch(citiesIds, destination);
double [] distancesFromSource = g.distancesFrom(sourceIndex);
int destinationDistance = (int)distancesFromSource[destinationIndex];
System.out.println(destinationDistance);
Comment puis-je éviter ce NullPointerException?
The complete code:
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package tshpath;
import java.io.*;
import java.util.*;
class Graph {
private double [][]edges;
/*el argumento es el número de vértices en este grafo*/
public Graph(int vertices){
edges = new double [vertices][vertices];
}
/*añade una arista de peso 1 a partir de i hasta j*/
public void addEdge(int i, int j){
edges[i][j]=1;
}
/*añade aristas de peso 1 de i hasta j y de j hasta i*/
public void addUndirectedEdge (int i, int j){
edges[i][j]=1;
edges[j][i]=1;
}
/*retorna el costo de la arista de i y j*/
public double getEdge(int i, int j){
return edges[i][j];
}
/*retorna true si hay una arista entre i y j*/
public boolean hasEdge (int i, int j){
return edges[i][j] !=0.0;
}
/*fija el peso de la arista entre i y j*/
public void setEdge (int i, int j, double weight){
edges [i][j] = weight;
}
/*fija el peso de la arista entre i y j y entre j e i*/
public void setUndirectedEdge (int i, int j, double weight){
edges[i][j] = weight;
edges[j][i] = weight;
}
/*retorna el número de vértices en este grafo*/
public int size() {
return edges.length;
}
/*retorna una lista de los vecinos del vértice i*/
public List <Integer> neighbors (int i){
List <Integer> result = new ArrayList<Integer>();
for (int j=0; j<size();j++){
if (hasEdge(i,j)){
result.add(j);
}
}
return result;
}
/*retorna 0 si i y j son idénticos, retorna infinito si no hay arista entre ellos o si
* el peso entre las aristas si hay uno*/
public double getCost(int i , int j){
if (i==j){
return 0.0;
}
if (edges[i][j]==0.0){
return Double.POSITIVE_INFINITY;
}
return edges[i][j];
}
/*dijkstra, retorna el índice del elemento más pequeño de distances, ignorando
* aquellos en visited*/
protected int cheapest (double [] distances, boolean [] visited){
int best =-1;
for (int i=0; i<size(); i++){
if (!visited[i]
&& ((best < 0) || (distances[i] < distances[best]))) {
best =i;
}
}
return best;
}
public double [] distancesFrom (int source){
double [] result = new double[size()];
java.util.Arrays.fill(result, Double.POSITIVE_INFINITY);
result [source]=0;
boolean []visited = new boolean [size()];
for (int i =0; i<size();i++){
int vertex = cheapest (result,visited);
visited [vertex]=true;
for (int j =0; j<size();j++){
result [j] = Math.min(result[j], result[vertex]+getCost(vertex,j));
}
}
return result;
}
/*test Graph*/
/*public static void main(String args[]){
Graph g = new Graph(5);
g.setEdge(1,2,1);
g.setEdge(1,3,3);
g.setEdge(2,1,1);
g.setEdge(2,3,1);
g.setEdge(2,4,4);
g.setEdge(3,1,3);
g.setEdge(3,2,1);
g.setEdge(3,4,1);
g.setEdge(4,2,4);
g.setEdge(4,3,1);
double [] distancesFrom1 = g.distancesFrom(1);
double [] distancesFrom2 = g.distancesFrom(2);
System.out.println((int)distancesFrom1[4]);
System.out.println((int)distancesFrom2[4]);
}*/
}
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws Exception {
// TODO code application logic here
BufferedReader r = new BufferedReader (new FileReader(new File("C:\\Documents and Settings\\Administrator\\My Documents\\NetBeansProjects\\TSHPATH\\src\\tshpath\\TSHPATHInput.txt")));
//BufferedReader r1 = new BufferedReader (new InputStreamReader(System.in));
String line = r.readLine();
// System.out.println(line); //Linea de prueba
int s = Integer.parseInt(line);
for (int testIndex=0; testIndex<s; testIndex++){
String [] citiesIds = new String[10000];
line = r.readLine();
//System.out.println(line); //Linea de prueba
int n = Integer.parseInt(line);
int graphSize = n +1; // por el problema de indexación desde 0 en el arreglo
Graph g = new Graph (graphSize);
for (int cityIndex=0; cityIndex<n;cityIndex++){
line = r.readLine();
// System.out.println(line); //Linea de prueba
String NAME = line;
int auxCityIndex = cityIndex +1; // para mantener la consistencia en la indexación
citiesIds[auxCityIndex] = NAME;
line = r.readLine();
// System.out.println(line); //Linea de prueba
int p = Integer.parseInt(line);
for (int neighborIndex=0;neighborIndex<p;neighborIndex++){
line = r.readLine();
// System.out.println(line); //Linea de prueba
String [] brokenLine = line.split(" ");
int cityToConnect = Integer.parseInt(brokenLine[0]);
int weightOfConnection = Integer.parseInt(brokenLine[1]);
g.setEdge(auxCityIndex,cityToConnect, weightOfConnection);
}
}
line = r.readLine();
//System.out.println(line); //Linea de prueba
int routesToFind = Integer.parseInt(line);
for (int routesIndex=0; routesIndex<routesToFind; routesIndex++){
line = r.readLine();
// System.out.println(line); //Linea de prueba
String [] cityNames = line.split(" ");
String source = cityNames[0];
String destination = cityNames[1];
int sourceIndex = Arrays.binarySearch(citiesIds,source);
int destinationIndex= Arrays.binarySearch(citiesIds, destination);
double [] distancesFromSource = g.distancesFrom(sourceIndex);
int destinationDistance = (int)distancesFromSource[destinationIndex];
System.out.println(destinationDistance);
}
}
}
}
le fichier d'entrée:
1
4
gdansk
2
2 1
3 3
bydgoszcz
3
1 1
3 1
4 4
torun
3
1 3
2 1
4 1
warszawa
2
2 4
3 1
2
gdansk warszawa
bydgoszcz warszawa
Ah, beaucoup mieux. – CPerkins