Java實現最短路徑算法(Dijkstra 算法)


參考:

https://zhuanlan.zhihu.com/p/129373740

《數據結構與算法-python描述》作者:裘宗燕

 

 以上是原圖,求V1到其余所有節點的最短路徑。參考了裘宗燕教授的數據結構與算法

並未完全理解其精髓,暫且記錄,后面再慢慢理解

@Data
@AllArgsConstructor
public class WeightGraph {
    private List<String> nodes;
    private List<WeightEdge<String, String, Integer>> edgeList;


    public static void main(String[] args) {
        WeightGraph weightGraph = buildGraph();
        Map<String, Path<String, Integer>> shortest = findShortest(weightGraph);
        for (Map.Entry<String, Path<String, Integer>> entry : shortest.entrySet()) {
            System.out.println(entry.getKey() + ":" + entry.getValue().getLen());
        }
    }


    /**
     * 初始化圖結構
     *
     * @return
     */
    public static WeightGraph buildGraph() {
        List<String> node = Lists.newArrayList("v1", "v2", "v3", "v4", "v5", "v6");
        List<WeightEdge<String, String, Integer>> edge = Lists.newArrayList();
        edge.add(new WeightEdge<>("v1", "v2", 10));
        edge.add(new WeightEdge<>("v2", "v3", 7));
        edge.add(new WeightEdge<>("v4", "v3", 4));
        edge.add(new WeightEdge<>("v4", "v5", 7));
        edge.add(new WeightEdge<>("v6", "v5", 1));
        edge.add(new WeightEdge<>("v1", "v6", 3));
        edge.add(new WeightEdge<>("v6", "v2", 2));
        edge.add(new WeightEdge<>("v4", "v1", 3));
        edge.add(new WeightEdge<>("v2", "v4", 5));
        edge.add(new WeightEdge<>("v6", "v4", 6));

        return new WeightGraph(node, edge);
    }


    public static Map<String, Path<String, Integer>> findShortest(WeightGraph weightGraph) {
        List<WeightEdge<String, String, Integer>> edgeList = weightGraph.getEdgeList();
        List<String> nodes = weightGraph.getNodes();
        //以路徑邊長做key的優先級隊列
        PriorityQueue<Triple<Integer, String, String>> queue = new PriorityQueue<>(Comparator.comparing(Triple::getLen));
        //初始化隊列
        //表示原點v1,經中間節點v1,到v1的距離是0
        queue.add(new Triple<>(0, "v1", "v1"));
        //結果map
        Map<String, Path<String, Integer>> paths = new HashMap<>();
        //paths初始為null,用於存儲最終結果
        int size = nodes.size();
        for (String node : nodes) {
            paths.put(node, null);
        }

        int count = 0;
        while (count < size && !queue.isEmpty()) {
            //長度、中介、終點
            Triple<Integer, String, String> triple = queue.remove();
            String end = triple.getEnd();
            //如果end已經找到,則跳過本次循環
            if (paths.get(end) != null) {
                continue;
            }
            //如果end還為找到最短路徑
            Integer len = triple.getLen();
            //因為從優先隊列頭部取出,因此<end,len>就是當前已知的最短路徑
            //第一次執行("v1",new Path("v1",0)),v1經v1到v1的最短距離是0
            paths.put(end, new Path<>(triple.getMiddle(), len));
            //遍歷所有以新得到的最短路徑節點做start的邊,擴展可達的邊界
            for (WeightEdge<String, String, Integer> edge : edgeList) {
                if (edge.getStart().equals(end)) {
                    String newEnd = edge.getEnd();
                    Integer w = edge.getWeight();
                    //如果新目標還未確定最短路徑,則將end做中介,原end的len+weight作為新的len,並加入優先queue
                    //優先隊列保證經過這輪循環,得到了最新的最短路徑節點
                    if (paths.get(newEnd) == null) {
                        queue.add(new Triple<>(len + w, end, newEnd));
                    }
                }
            }
            count += 1;
        }
        return paths;
    }
}

/**
 * 帶權重的邊
 *
 * @param <S>
 * @param <E>
 * @param <W>
 */
@Data
@AllArgsConstructor
@NoArgsConstructor
class WeightEdge<S, E, W> {
    private S start;
    private E end;
    private W weight;


}

/**
 * 優先隊列元素三元組,主要用於獲取目標節點的最短路徑
 *
 * @param <L> length
 * @param <M> 中介點
 * @param <E> 目標點
 */
@Data
@AllArgsConstructor
class Triple<L, M, E> {
    private L len;
    private M middle;
    private E end;
}

/**
 * paths的元素格式
 *
 * @param <M> 中介點
 * @param <L> 最終最短路徑
 */
@Data
@AllArgsConstructor
class Path<M, L> {
    private M middle;
    private L len;
}

輸出:

v6:3
v1:0
v2:5
v3:12
v4:9
v5:4

 


免責聲明!

本站轉載的文章為個人學習借鑒使用,本站對版權不負任何法律責任。如果侵犯了您的隱私權益,請聯系本站郵箱yoyou2525@163.com刪除。



 
粵ICP備18138465號   © 2018-2025 CODEPRJ.COM