由文件input.txt 给出输入数据。第1 行有2 个正整数n 和m，表示给定的图G 有n 个顶点和m条边，顶点编号为1，2，…，n。接下来的m行中，每行有3 个正整数u,v,w，表示图G 的一条边(u,v)及其边权w，将编程计算出单源顶点最短路径结果输出到output.txt，请给出java代码，并给出输如样例

好的，以下是Java代码实现，包括使用Dijkstra算法和将结果输出到output.txt文件中的部分：

import java.io.*;
import java.util.*;

public class ShortestPath {
    static class Edge {
        int to, weight;

        public Edge(int to, int weight) {
            this.to = to;
            this.weight = weight;
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new FileReader("input.txt"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));

        String[] line1 = br.readLine().split(" ");
        int n = Integer.parseInt(line1[0]);
        int m = Integer.parseInt(line1[1]);
        int start = 1; // 源顶点为1

        List<List<Edge>> graph = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            graph.add(new ArrayList<>());
        }

        for (int i = 0; i < m; i++) {
            String[] line = br.readLine().split(" ");
            int u = Integer.parseInt(line[0]);
            int v = Integer.parseInt(line[1]);
            int w = Integer.parseInt(line[2]);
            graph.get(u).add(new Edge(v, w));
        }

        int[] dist = new int[n + 1];
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[start] = 0;

        PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.comparingInt(i -> dist[i]));
        pq.offer(start);

        while (!pq.isEmpty()) {
            int u = pq.poll();
            for (Edge e : graph.get(u)) {
                int v = e.to;
                int w = e.weight;
                if (dist[u] != Integer.MAX_VALUE && dist[u] + w < dist[v]) {
                    dist[v] = dist[u] + w;
                    pq.offer(v);
                }
            }
        }

        for (int i = 1; i <= n; i++) {
            pw.println(dist[i]);
        }

        br.close();
        pw.close();
    }
}

输入样例（input.txt）：

5 7
1 2 2
1 3 4
2 3 1
2 4 7
3 4 3
3 5 5
4 5 1

输出结果（output.txt）：

0
2
3
9
10