2. 给定带权有向图G =（V，E），其中每条边的权是非负实数。另外，还给定V中的一个顶点，称为源。现在要计算从源到其他各顶点的最短路长度。这里路的长度是指路上各边权之和。的java程序

可以使用Dijkstra算法或者Bellman-Ford算法来解决该问题。以下是使用Dijkstra算法的Java程序示例：

import java.util.*;

public class ShortestPath {
    private int[] dist;
    private boolean[] visited;
    private PriorityQueue<Node> pq;
    private List<List<Node>> adjList;

    private static class Node implements Comparable<Node> {
        int vertex;
        int weight;

        Node(int vertex, int weight) {
            this.vertex = vertex;
            this.weight = weight;
        }

        public int compareTo(Node other) {
            return Integer.compare(weight, other.weight);
        }
    }

    public ShortestPath(int n) {
        dist = new int[n];
        visited = new boolean[n];
        pq = new PriorityQueue<Node>();
        adjList = new ArrayList<List<Node>>();

        for (int i = 0; i < n; i++) {
            dist[i] = Integer.MAX_VALUE;
            visited[i] = false;
            adjList.add(new ArrayList<Node>());
        }
    }

    public void addEdge(int u, int v, int weight) {
        adjList.get(u).add(new Node(v, weight));
    }

    public void dijkstra(int source) {
        dist[source] = 0;
        pq.offer(new Node(source, 0));

        while (!pq.isEmpty()) {
            Node node = pq.poll();
            int u = node.vertex;

            if (visited[u]) {
                continue;
            }

            visited[u] = true;

            for (Node edge : adjList.get(u)) {
                int v = edge.vertex;
                int weight = edge.weight;

                if (!visited[v] && dist[u] + weight < dist[v]) {
                    dist[v] = dist[u] + weight;
                    pq.offer(new Node(v, dist[v]));
                }
            }
        }
    }

    public static void main(String[] args) {
        ShortestPath sp = new ShortestPath(5);

        sp.addEdge(0, 1, 10);
        sp.addEdge(0, 4, 5);
        sp.addEdge(1, 2, 1);
        sp.addEdge(1, 4, 2);
        sp.addEdge(2, 3, 4);
        sp.addEdge(3, 2, 6);
        sp.addEdge(3, 0, 7);
        sp.addEdge(4, 1, 3);
        sp.addEdge(4, 2, 9);
        sp.addEdge(4, 3, 2);

        sp.dijkstra(0);

        for (int i = 0; i < 5; i++) {
            System.out.println("Shortest distance from source to " + i + " is " + sp.dist[i]);
        }
    }
}

输出结果如下：

Shortest distance from source to 0 is 0
Shortest distance from source to 1 is 8
Shortest distance from source to 2 is 9
Shortest distance from source to 3 is 7
Shortest distance from source to 4 is 5

该程序使用了优先队列来维护未访问的节点，并通过计算从源节点到当前节点的距离来更新最短距离数组。