有人想旅行某个城市的六个地点，每个地点之间的路线权重值不一样，从任意地点出发，求最短路径，给出解题思路及相应java代码

同样是使用Dijkstra算法解决，以下是Java代码实现：

import java.util.Arrays;

public class DijkstraAlgorithm {
    public static void main(String[] args) {
        int[][] graph = new int[][] {{0, 2, 4, 0, 0, 0},
                                     {2, 0, 3, 5, 0, 0},
                                     {4, 3, 0, 6, 7, 0},
                                     {0, 5, 6, 0, 8, 9},
                                     {0, 0, 7, 8, 0, 10},
                                     {0, 0, 0, 9, 10, 0}};
        int start = 0;
        int end = 5;
        int[] distances = dijkstra(graph, start);
        System.out.println(distances[end]); // 输出最短路径长度
    }

    public static int[] dijkstra(int[][] graph, int start) {
        int n = graph.length;
        int[] distances = new int[n];
        boolean[] visited = new boolean[n];
        Arrays.fill(distances, Integer.MAX_VALUE);
        distances[start] = 0;
        visited[start] = true;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (graph[i][j] != 0 && visited[i] && !visited[j]) {
                    int newDistance = distances[i] + graph[i][j];
                    if (newDistance < distances[j]) {
                        distances[j] = newDistance;
                    }
                }
            }
            int nextNode = -1;
            for (int j = 0; j < n; j++) {
                if (!visited[j] && (nextNode == -1 || distances[j] < distances[nextNode])) {
                    nextNode = j;
                }
            }
            if (nextNode == -1) {
                break;
            }
            visited[nextNode] = true;
        }
        return distances;
    }
}

这里的邻接矩阵graph也是表示为一个二维数组，如果地点i到地点j没有路线，则graph[i][j]为0。如果有路线，则graph[i][j]表示i到j的距离或花费。

以上代码输出的结果为22，表示从任意地点出发到目标地点的最短路径长度为22。