数据结构与算法：图

图

1) 概念

图是由顶点（vertex）和边（edge）组成的数据结构，例如

该图有四个顶点：A、B、C、D 以及四条有向边，有向图中，边是单向的

有向 vs 无向

如果是无向图，那么边是双向的，下面是一个无向图的例子

度

度是指与该顶点相邻的边的数量

例如上图中

• A、B、C、E、F 这几个顶点度数为 2
• D 顶点度数为 4

有向图中，细分为入度和出度，参见下图

• A (2 out / 0 in)
• B、C、E (1 out / 1 in)
• D (2 out / 2 in)
• F (0 out / 2 in)

权

边可以有权重，代表从源顶点到目标顶点的距离、费用、时间或其他度量。

路径

路径被定义为从一个顶点到另一个顶点的一系列连续边，例如上图中【北京】到【上海】有多条路径

• 北京 - 上海
• 北京 - 武汉 - 上海

路径长度

• 不考虑权重，长度就是边的数量
• 考虑权重，一般就是权重累加

环

在有向图中，从一个顶点开始，可以通过若干条有向边返回到该顶点，那么就形成了一个环

图的连通性

如果两个顶点之间存在路径，则这两个顶点是连通的，所有顶点都连通，则该图被称之为连通图，若子图连通，则称为连通分量

2) 图的表示

比如说，下面的图

用邻接矩阵可以表示为：

  A B C D
A 0 1 1 0
B 1 0 0 1 
C 1 0 0 1
D 0 1 1 0

用邻接表可以表示为：

A -> B -> C
B -> A -> D
C -> A -> D
D -> B -> C

有向图的例子

  A B C D
A 0 1 1 0
B 0 0 0 1
C 0 0 0 1
D 0 0 0 0

A - B - C
B - D
C - D
D - empty

3) Java 表示

顶点

public class Vertex {
    String name;
    List<Edge> edges;

    // 拓扑排序相关
    int inDegree;
    int status; // 状态 0-未访问 1-访问中 2-访问过，用在拓扑排序

    // dfs, bfs 相关
    boolean visited;

    // 求解最短距离相关
    private static final int INF = Integer.MAX_VALUE;
    int dist = INF;
    Vertex prev = null;

    @Override
    public String toString() {
        return this.name;
    }
}

边

public class Edge {

    Vertex linked;
    int weight;

    public Edge(Vertex linked) {
        this(linked, 1);
    }

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

4) DFS

public class Dfs {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");

        v1.edges = List.of(new Edge(v3), new Edge(v2), new Edge(v6));
        v2.edges = List.of(new Edge(v4));
        v3.edges = List.of(new Edge(v4), new Edge(v6));
        v4.edges = List.of(new Edge(v5));
        v5.edges = List.of();
        v6.edges = List.of(new Edge(v5));

        dfs1(v1);
    }

    private static void dfs2(Vertex v) {
        LinkedList<Vertex> stack = new LinkedList<>();
        stack.push(v);
        while (!stack.isEmpty()) {
            Vertex pop = stack.pop();
            pop.visited = true;
            System.out.println(pop.name);
            for (Edge edge : pop.edges) {
                if (!edge.linked.visited) {
                    stack.push(edge.linked);
                }
            }
        }
    }

    private static void dfs1(Vertex v) {
        v.visited = true;
        System.out.println(v.name);
        for (Edge edge : v.edges) {
            if (!edge.linked.visited) {
                dfs(edge.linked);
            }
        }
    }
}

5) BFS

public class Bfs {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");

        v1.edges = List.of(new Edge(v3), new Edge(v2), new Edge(v6));
        v2.edges = List.of(new Edge(v4));
        v3.edges = List.of(new Edge(v4), new Edge(v6));
        v4.edges = List.of(new Edge(v5));
        v5.edges = List.of();
        v6.edges = List.of(new Edge(v5));

        bfs(v1);
    }

    private static void bfs(Vertex v) {
        LinkedList<Vertex> queue = new LinkedList<>();
        v.visited = true;
        queue.offer(v);
        while (!queue.isEmpty()) {
            Vertex poll = queue.poll();
            System.out.println(poll.name);
            for (Edge edge : poll.edges) {
                if (!edge.linked.visited) {
                    edge.linked.visited = true;
                    queue.offer(edge.linked);
                }
            }
        }
    }
}

6) 拓扑排序

public class TopologicalSort {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("网页基础");
        Vertex v2 = new Vertex("Java基础");
        Vertex v3 = new Vertex("JavaWeb");
        Vertex v4 = new Vertex("Spring框架");
        Vertex v5 = new Vertex("微服务框架");
        Vertex v6 = new Vertex("数据库");
        Vertex v7 = new Vertex("实战项目");

        v1.edges = java.util.List.of(new Edge(v3)); // +1
        v2.edges = java.util.List.of(new Edge(v3)); // +1
        v3.edges = java.util.List.of(new Edge(v4));
        v6.edges = java.util.List.of(new Edge(v4));
        v4.edges = java.util.List.of(new Edge(v5));
        v5.edges = java.util.List.of(new Edge(v7));
        v7.edges = java.util.List.of();
        List<Vertex> graph = java.util.List.of(v1,v2,v3,v4,v5,v6,v7);
        for (Vertex vertex : graph) {
            for (Edge edge : vertex.edges) {
                edge.linked.inDegree += 1;
            }
        }

        List<Vertex> result = new ArrayList<>();
        Stack<Vertex> stack = new Stack<>();
        for (Vertex vertex : graph) {
            if(vertex.inDegree == 0 ){
                stack.push(vertex);
            }
        }
        while(!stack.isEmpty()){
            Vertex pop = stack.pop();
            result.add(pop);
            for (Edge edge : pop.edges) {
                edge.linked.inDegree--;
                if(edge.linked.inDegree == 0 && !edge.linked.visited){
                    stack.add(edge.linked);
                    edge.linked.visited = true;
                }
            }
        }
        if(result.size() != graph.size()){
            System.out.println("发现环");
        }
        for (Vertex vertex : result) {
            System.out.println(vertex);
        }

    }

7) 最短路径

Dijkstra

算法描述：

1. 将所有顶点标记为未访问。创建一个未访问顶点的集合。
2. 为每个顶点分配一个临时距离值
   • 对于我们的初始顶点，将其设置为零
   • 对于所有其他顶点，将其设置为无穷大。
3. 每次选择最小临时距离的未访问顶点，作为新的当前顶点
4. 对于当前顶点，遍历其所有未访问的邻居，并更新它们的临时距离为更小
   • 例如，1->6 的距离是 14，而1->3->6 的距离是11。这时将距离更新为 11
   • 否则，将保留上次距离值
5. 当前顶点的邻居处理完成后，把它从未访问集合中删除

public class Dijkstra {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");

        v1.edges = List.of(new Edge(v3, 9), new Edge(v2, 7), new Edge(v6, 14));
        v2.edges = List.of(new Edge(v4, 15));
        v3.edges = List.of(new Edge(v4, 11), new Edge(v6, 2));
        v4.edges = List.of(new Edge(v5, 6));
        v5.edges = List.of();
        v6.edges = List.of(new Edge(v5, 9));

        List<Vertex> graph = List.of(v1, v2, v3, v4, v5, v6);

        dijkstra(graph, v1);
    }

    private static void dijkstra(List<Vertex> graph, Vertex source) {
        ArrayList<Vertex> list = new ArrayList<>(graph);
        source.dist = 0;

        while (!list.isEmpty()) {
            // 3. 选取当前顶点
            Vertex curr = chooseMinDistVertex(list);
            // 4. 更新当前顶点邻居距离
            updateNeighboursDist(curr, list);
            // 5. 移除当前顶点
            list.remove(curr);
        }

        for (Vertex v : graph) {
            System.out.println(v.name + " " + v.dist);
        }
    }

    private static void updateNeighboursDist(Vertex curr, ArrayList<Vertex> list) {
        for (Edge edge : curr.edges) {
            Vertex n = edge.linked;
            if (list.contains(n)) {
                int dist = curr.dist + edge.weight;
                if (dist < n.dist) {
                    n.dist = dist;
                }
            }
        }
    }

    private static Vertex chooseMinDistVertex(ArrayList<Vertex> list) {
        Vertex min = list.get(0);
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i).dist < min.dist) {
                min = list.get(i);
            }
        }
        return min;
    }

}

改进 - 优先级队列

1. 创建一个优先级队列，放入所有顶点（队列大小会达到边的数量）
2. 为每个顶点分配一个临时距离值
   • 对于我们的初始顶点，将其设置为零
   • 对于所有其他顶点，将其设置为无穷大。
3. 每次选择最小临时距离的未访问顶点，作为新的当前顶点
4. 对于当前顶点，遍历其所有未访问的邻居，并更新它们的临时距离为更小，若距离更新需加入队列
   • 例如，1->6 的距离是 14，而1->3->6 的距离是11。这时将距离更新为 11
   • 否则，将保留上次距离值
5. 当前顶点的邻居处理完成后，把它从队列中删除

public class DijkstraPriorityQueue {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");

        v1.edges = List.of(new Edge(v3, 9), new Edge(v2, 7), new Edge(v6, 14));
        v2.edges = List.of(new Edge(v4, 15));
        v3.edges = List.of(new Edge(v4, 11), new Edge(v6, 2));
        v4.edges = List.of(new Edge(v5, 6));
        v5.edges = List.of();
        v6.edges = List.of(new Edge(v5, 9));

        List<Vertex> graph = List.of(v1, v2, v3, v4, v5, v6);

        dijkstra(graph, v1);
    }

    private static void dijkstra(List<Vertex> graph, Vertex source) {
        PriorityQueue<Vertex> queue = new PriorityQueue<>(Comparator.comparingInt(v -> v.dist));
        source.dist = 0;
        for (Vertex v : graph) {
            queue.offer(v);
        }

        while (!queue.isEmpty()) {
            System.out.println(queue);
            // 3. 选取当前顶点
            Vertex curr = queue.poll();
            // 4. 更新当前顶点邻居距离
            if(!curr.visited) {
                updateNeighboursDist(curr, queue);
                curr.visited = true;
            }
            // 5. 移除当前顶点
        }

        for (Vertex v : graph) {
            System.out.println(v.name + " " + v.dist + " " + (v.prev != null ? v.prev.name : "null"));
        }
    }

    private static void updateNeighboursDist(Vertex curr, PriorityQueue<Vertex> queue) {
        for (Edge edge : curr.edges) {
            Vertex n = edge.linked;
            if (!n.visited) {
                int dist = curr.dist + edge.weight;
                if (dist < n.dist) {
                    n.dist = dist;
                    n.prev = curr;
                    queue.remove(n); // 先删除再添加才能改变优先级
                    queue.offer(n);
                }
            }
        }
    }

}

问题

按照 Dijkstra 算法，得出

• v1 -> v2 最短距离2
• v1 -> v3 最短距离1
• v1 -> v4 最短距离2

事实应当是

• v1 -> v2 最短距离2
• v1 -> v3 最短距离0
• v1 -> v4 最短距离1

Bellman-Ford

public class BellmanFord {
    public static void main(String[] args) {
        // 正常情况
        /*Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");

        v1.edges = List.of(new Edge(v3, 9), new Edge(v2, 7), new Edge(v6, 14));
        v2.edges = List.of(new Edge(v4, 15));
        v3.edges = List.of(new Edge(v4, 11), new Edge(v6, 2));
        v4.edges = List.of(new Edge(v5, 6));
        v5.edges = List.of();
        v6.edges = List.of(new Edge(v5, 9));

        List<Vertex> graph = List.of(v4, v5, v6, v1, v2, v3);*/

        // 负边情况
        /*Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");

        v1.edges = List.of(new Edge(v2, 2), new Edge(v3, 1));
        v2.edges = List.of(new Edge(v3, -2));
        v3.edges = List.of(new Edge(v4, 1));
        v4.edges = List.of();
        List<Vertex> graph = List.of(v1, v2, v3, v4);*/

        // 负环情况
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");

        v1.edges = List.of(new Edge(v2, 2));
        v2.edges = List.of(new Edge(v3, -4));
        v3.edges = List.of(new Edge(v4, 1), new Edge(v1, 1));
        v4.edges = List.of();
        List<Vertex> graph = List.of(v1, v2, v3, v4);

        bellmanFord(graph, v1);
    }

    private static void bellmanFord(List<Vertex> graph, Vertex source) {
        source.dist = 0;
        int size = graph.size();
        // 1. 进行 顶点个数 - 1 轮处理
        for (int i = 0; i < size - 1; i++) {
            // 2. 遍历所有的边
            for (Vertex s : graph) {
                for (Edge edge : s.edges) {
                    // 3. 处理每一条边
                    Vertex e = edge.linked;
                    if (s.dist != Integer.MAX_VALUE && s.dist + edge.weight < e.dist) {
                        e.dist = s.dist + edge.weight;
                        e.prev = s;
                    }
                }
            }
        }
        for (Vertex v : graph) {
            System.out.println(v + " " + (v.prev != null ? v.prev.name : "null"));
        }
    }
}

负环

如果在【顶点-1】轮处理完成后，还能继续找到更短距离，表示发现了负环

Floyd-Warshall

public class FloydWarshall {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");

        v1.edges = List.of(new Edge(v3, -2));
        v2.edges = List.of(new Edge(v1, 4), new Edge(v3, 3));
        v3.edges = List.of(new Edge(v4, 2));
        v4.edges = List.of(new Edge(v2, -1));
        List<Vertex> graph = List.of(v1, v2, v3, v4);

        /*
                直接连通
                v1  v2  v3  v4
            v1  0   ∞   -2  ∞
            v2  4   0   3   ∞
            v3  ∞   ∞   0   2
            v4  ∞   -1  ∞   0

                k=0 借助v1到达其它顶点
                v1  v2  v3  v4
            v1  0   ∞   -2  ∞
            v2  4   0   2   ∞
            v3  ∞   ∞   0   2
            v4  ∞   -1  ∞   0

                k=1 借助v2到达其它顶点
                v1  v2  v3  v4
            v1  0   ∞   -2  ∞
            v2  4   0   2   ∞
            v3  ∞   ∞   0   2
            v4  3   -1  1   0

                k=2 借助v3到达其它顶点
                v1  v2  v3  v4
            v1  0   ∞   -2  0
            v2  4   0   2   4
            v3  ∞   ∞   0   2
            v4  3   -1  1   0

                k=3 借助v4到达其它顶点
                v1  v2  v3  v4
            v1  0   -1   -2  0
            v2  4   0   2   4
            v3  5   1   0   2
            v4  3   -1  1   0
         */
        floydWarshall(graph);
    }

    private static void floydWarshall(List<Vertex> graph){
        int size = graph.size();
        int[][] dist = new int[size][size];
        for (int i = 0; i < size; i++) {  // 初始化
            Vertex vertex = graph.get(i);
            Map<Vertex, Integer> collect = vertex.edges.stream().collect(Collectors.toMap(v -> v.linked, v -> v.weight));
            for (int i1 = 0; i1 < size; i1++) {
                if(i == i1){
                    dist[i][i1] = 0;
                    continue;
            }
                dist[i][i1] = collect.getOrDefault(graph.get(i1),Integer.MAX_VALUE);
        }
        }
        for (int k = 0; k < size; k++) {
            for (int j = 0; j < size; j++) {
                int dist1;
                if((dist1 = dist[j][k]) < Integer.MAX_VALUE){
                    for (int i = 0; i < size; i++) {
                        int dist2;
                        if((dist2 = dist[k][i]) != Integer.MAX_VALUE)
                            dist[j][i] = Integer.min(dist1 + dist2,dist[j][i]);
                    }
                }
            }
        }

        for (int[] ints : dist) {
            for (int anInt : ints) {
                System.out.print(anInt + " ");
            }
            System.out.println();
        }
    }
}

负环

如果在 3 层循环结束后，在 dist 数组的对角线处（i==j 处）发现了负数，表示出现了负环

8) 最小生成树

图的最小生成树是一个子图，它是连通的，包含图中所有的顶点，并且所有边的权重之和最小。在最小生成树中，没有任何一条边可以被其他边替换而使得总权重变小。也就是说，最小生成树是图的所有生成树中，边的权值总和最小的生成树。

解决最小生成树问题的常用算法有Prim算法和Kruskal算法。Prim算法从一个顶点开始，每次都添加一条与当前子图连接的权重最小的边，直到所有顶点都被包含在子图中。Kruskal算法则是从所有的边开始，每次都添加一条当前所有边中权重最小的边，但需要保证添加的边不会形成环，直到所有顶点都被连接。

Prim

public class Prim {
    public static void main(String[] args) {
        Vertex v1 = new Vertex("v1");
        Vertex v2 = new Vertex("v2");
        Vertex v3 = new Vertex("v3");
        Vertex v4 = new Vertex("v4");
        Vertex v5 = new Vertex("v5");
        Vertex v6 = new Vertex("v6");
        Vertex v7 = new Vertex("v7");

        v1.edges = List.of(new Edge(v2, 2), new Edge(v3, 4), new Edge(v4, 1));
        v2.edges = List.of(new Edge(v1, 2), new Edge(v4, 3), new Edge(v5, 10));
        v3.edges = List.of(new Edge(v1, 4), new Edge(v4, 2), new Edge(v6, 5));
        v4.edges = List.of(new Edge(v1, 1), new Edge(v2, 3), new Edge(v3, 2),
                new Edge(v5, 7), new Edge(v6, 8), new Edge(v7, 4));
        v5.edges = List.of(new Edge(v2, 10), new Edge(v4, 7), new Edge(v7, 6));
        v6.edges = List.of(new Edge(v3, 5), new Edge(v4, 8), new Edge(v7, 1));
        v7.edges = List.of(new Edge(v4, 4), new Edge(v5, 6), new Edge(v6, 1));

        List<Vertex> graph = List.of(v1, v2, v3, v4, v5, v6, v7);

        prim(graph, v1);

    }

    static void prim(List<Vertex> graph, Vertex source) {
        ArrayList<Vertex> list = new ArrayList<>(graph);
        source.dist = 0;

        while (!list.isEmpty()) {
            Vertex min = chooseMinDistVertex(list);
            updateNeighboursDist(min);
            list.remove(min);
            min.visited = true;
            System.out.println("---------------");
            for (Vertex v : graph) {
                System.out.println(v);
            }
        }


    }

    private static void updateNeighboursDist(Vertex curr) {
        for (Edge edge : curr.edges) {
            Vertex n = edge.linked;
            if (!n.visited) {
                int dist = edge.weight;
                if (dist < n.dist) {
                    n.dist = dist;
                    n.prev = curr;
                }
            }
        }
    }

    private static Vertex chooseMinDistVertex(ArrayList<Vertex> list) {
        Vertex min = list.get(0);
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i).dist < min.dist) {
                min = list.get(i);
            }
        }
        return min;
    }
}

Kruskal

  private static void kruskal(List<Vertex> graph,Vertex v1){
    List<Edge> edges = new ArrayList<>();
    List<Vertex> pre = new ArrayList<>();
    for (Vertex vertex : graph) {
        for (Edge edge : vertex.edges) {
            edges.add(edge);
            pre.add(vertex);
        }
    }
    for (int i = 0; i < edges.size(); i++) {
        Edge minEdge = edges.get(i);
        int min = minEdge.weight;
        for (int j = i + 1 ; j < edges.size(); j++) {
            Edge e = edges.get(j);
            if(minEdge.weight > e.weight){
                edges.set(j,minEdge);
                minEdge = e;
                edges.set(i,e);
                Vertex v = pre.get(i);
                pre.set(i,pre.get(j));
                pre.set(j,v);
            }
        }
    }
    List<Vertex> used = new ArrayList<>();
    for (int i = 0; i < edges.size(); i++) {
        Vertex v = pre.get(i);
        Vertex e = edges.get(i).linked;
        if(!used.contains(v) || !used.contains(e)){
            System.out.println(v.name + " -> " + e.name);
            used.add(v);
            used.add(e);
        }
    }
}