有向图的基本算法-Java实现

有向图

有向图同无向图的区别为每条边带有方向，表明从一个顶点至另一个顶点可达。有向图的算法多依赖深度搜索算法。

本文主要介绍有向图的基本算法，涉及图的表示、可达性、检测环、图的遍历、拓扑排序以及强连通检测等算法。

1 定义有向图

采用邻接表结构存储边信息，同时提供reverse接口生成反向图，倒置每个边的方向，该接口在后续其他算法中会用到。

/**
 * 采用邻接表表示的有向图
 */
public class DiGraph {
    private final int V;
    private int E;
    private ArrayList<Integer>[] adj;
    public DiGraph(int V)
    {
        this.V = V;
        E = 0;
        adj = new ArrayList[V];
        for (int i = 0; i < V; i++) {
            adj[i] = new ArrayList<>();
        }
    }
    public DiGraph(Scanner scanner)
    {
        this(scanner.nextInt());
        int E = scanner.nextInt();
        for (int i = 0; i < E; i++) {
            int v = scanner.nextInt();
            int w = scanner.nextInt();
            addEdge(v, w);
        }
    }
    public void addEdge(int v, int w)
    {
        // 添加一条v指向w的边
        adj[v].add(w);
        E++;
    }
    /**
     * 返回有向图的反向图, 将每条边的方向反转
     */
    public DiGraph reverse()
    {
        DiGraph diGraph = new DiGraph(V);
        for (int v = 0; v < V; v++) {
            for (int w : adj[v]) {
                diGraph.addEdge(w, v);
            }
        }
        return diGraph;
    }
    public void show() {
        System.out.println("V: " + V);
        System.out.println("E: " + E);
        for (int i = 0; i < V; i++) {
            System.out.print(i + ": ");
            for (Integer integer : adj[i]) {
                System.out.print(integer + " ");
            }
            System.out.println();
        }
    }
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // 输入用例参加附录1
        DiGraph diGraph = new DiGraph(scanner);
        // 输入结果见附录2
        diGraph.show();
    }
}

2 有向图的可达性

有向图的可达性是指给定一个或一组顶点，判断是否可以到达图中其他顶点。垃圾清除常见算法“标记-清除”算法中，采用有向图的可达性算法

标记所有可以被访问的对象，然后在回收阶段，仅仅回收那些未被标记的对象。

/**
 * 基于深度优先的有向图可达性算法
 * 求出给定顶点或一组顶点，有向图中能到达的点
 */
public class DirectedDFS {
    private boolean[] marked;   // 标记每个顶点是否可到达
    public DirectedDFS(DiGraph G, int s)
    {
        marked = new boolean[G.V()];
        dfs(G, s);
    }
    public DirectedDFS(DiGraph G, Iterable<Integer> sources)
    {
        marked = new boolean[G.V()];
        for (int v : sources) {
            if(!marked[v]){
                dfs(G, v);
            }
        }
    }
    private void dfs(DiGraph G, int v)
    {
        marked[v] = true;
        for (int w : G.adj(v)) {
            if(!marked[w])
                dfs(G, w);
        }
    }
    public boolean marked(int v) { return marked[v]; }
    public static void main(String[] args) {
        // 输入用例参加附录1
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        // 输出结果参加附录3
        // 测试顶点2到达的点
        System.out.println("顶点2到达的点");
        DirectedDFS reachable = new DirectedDFS(diGraph, 2);
        for (int i = 0; i < diGraph.V(); i++)
            if(reachable.marked(i)) System.out.print(i + " ");
        System.out.println();
        // 测试一组点：1，2，6能够到达的点
        System.out.println("1，2，6能够到达的点");
        DirectedDFS reachable2 = new DirectedDFS(diGraph, Arrays.asList(1, 2, 6));
        for (int i = 0; i < diGraph.V(); i++)
            if(reachable2.marked(i)) System.out.print(i + " ");
        System.out.println();
    }
}

3 单点有向路径和单点最短有向路径

分别采用深度优先搜索和广度优先搜索实现

有向图的路径

/**
 * 单点有向路径，给定顶点v，确定对于图中任一点w;
 * 是否存在v到w的路径，并输出路径;
 * 注意，深度优先搜索的路径无法保证是最短路径
 */
public class DigraghDepthFirstPaths {
    // 标记点是否可达
    private boolean[] marked;
    // 记录到达点的那条边
    private int[] edge;
    private final int s;
    public DigraghDepthFirstPaths(DiGraph G, int s)
    {
        this.s = s;
        marked = new boolean[G.V()];
        edge = new int[G.V()];
        edge[s] = s;
        dfs(G, s);
    }
    private void dfs(DiGraph G, int v)
    {
        marked[v] = true;
        for (int w : G.adj(v)) {
            if(!marked[w]){
                edge[w] = v;
                dfs(G, w);
            }
        }
    }
    public boolean hasPathTo(int v){ return marked[v]; }
    public Stack<Integer> pathTo(int v)
    {
        Stack<Integer> paths = new Stack<>();
        for (int x=v; x!=s; x=edge[x]){
           paths.add(x);
        }
        paths.add(s);
        return paths;
    }
    public static void main(String[] args) {
        // 输入用例参加附录1
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        // 输出结果参加附录4
        // 构建顶点0到其他顶点的有向路径
        DigraghDepthFirstPaths depthFirstPaths = new DigraghDepthFirstPaths(diGraph, 0);
        System.out.print("顶点0可达的点: ");
        for (int i = 0; i < diGraph.V(); i++) {
            if (depthFirstPaths.hasPathTo(i)) System.out.print(i + " ");
        }
        System.out.println();
        // 是否存在有向路径
        if(depthFirstPaths.hasPathTo(12))
            System.out.println("0至12存在有向路径");
        else
            System.out.println("0至12不存在有向路径");
        // 顶点0到顶点3的一条有向路径
        System.out.print("0至3的一条有向路径: ");
        Stack<Integer> pathTo = depthFirstPaths.pathTo(3);
        while (!pathTo.isEmpty()){
            if (pathTo.size() == 1)
                System.out.print(pathTo.pop());
            else
                System.out.print(pathTo.pop() + " -> ");
        }
        System.out.println();
    }
}

有向图的最短路径，基于广度优先算法

/**
 * 基于广度优先搜索的单向路径算法；
 * 在此方法下，求得的路径为最短路径（忽略边权重）
 */
public class DigraphBreadthFirstPaths {
    private boolean[] marked;
    // 采用队列保持带访问的顶点
    private ArrayDeque<Integer> enqueue;
    private int[] edge;
    private final int s;
    public DigraphBreadthFirstPaths(DiGraph G, int s)
    {
        this.s = s;
        marked = new boolean[G.V()];
        edge = new int[G.V()];
        enqueue = new ArrayDeque<>();
        enqueue.add(s);
        bfs(G);
    }
    private void bfs(DiGraph G)
    {
        while (!enqueue.isEmpty())
        {
            int v = enqueue.poll();
            for (int w : G.adj(v)) {
                if(!marked[w]){
                    edge[w] = v;
                    marked[w] = true;
                    enqueue.add(w);
                }
            }
        }
    }
    public boolean hasPathTo(int v){ return marked[v]; }
    public Stack<Integer> pathTo(int v)
    {
        Stack<Integer> paths = new Stack<>();
        for (int x=v; x!=s; x=edge[x]){
            paths.add(x);
        }
        paths.add(s);
        return paths;
    }
    public static void main(String[] args) {
        // 输入用例参加附录1
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        // 输出结果参加附录5
        // 构建顶点0到其他顶点的有向路径
        DigraphBreadthFirstPaths breadthFirstPaths = new DigraphBreadthFirstPaths(diGraph, 0);
        System.out.print("顶点0可达的点: ");
        for (int i = 0; i < diGraph.V(); i++) {
            if (breadthFirstPaths.hasPathTo(i)) System.out.print(i + " ");
        }
        System.out.println();
        // 是否存在有向路径
        if(breadthFirstPaths.hasPathTo(12))
            System.out.println("0至12存在有向路径");
        else
            System.out.println("0至12不存在有向路径");
        // 顶点0到顶点3的最短路径
        System.out.print("0至3的一条有向路径: ");
        Stack<Integer> pathTo = breadthFirstPaths.pathTo(3);
        while (!pathTo.isEmpty()){
            if (pathTo.size() == 1)
                System.out.print(pathTo.pop());
            else
                System.out.print(pathTo.pop() + " -> ");
        }
        System.out.println();
    }
}

4 检测有向图的环

检测有向图是否包含环，检测图没有环是拓扑排序的前提条件。

多数情况下，需要知道有向图是否包含环，并且输出够成环的边。

/**
 * 基于深度优先搜索检测图中是否包含环
 */
public class DirectedCycle {
    private boolean[] onStack;
    private Stack<Integer> cycle;
    private int[] edge;
    private boolean[] marked;
    public DirectedCycle(DiGraph G)
    {
        onStack = new boolean[G.V()];
        edge = new int[G.V()];
        marked = new boolean[G.V()];
        for (int i = 0; i < G.V(); i++) {
            if(!marked[i])
                dfs(G, i);
        }
    }
    private void dfs(DiGraph G, int v)
    {
        onStack[v] = true;
        marked[v] = true;
        for (int w : G.adj(v)) {
            if (this.hasCycle()) return;
            else if (!marked[w]){
                edge[w] = v; dfs(G, w); }
            // onStack[w]为true表明，当前v节点是一条经过w的抵达，表明w -> v有路径
            // 由于v -> w有边，因此必为环
            else if(onStack[w]){
                cycle = new Stack<>();
                for (int x = v; x != w; x=edge[x])
                    cycle.push(x);
                cycle.push(w);
                cycle.push(v);
            }
        }
        onStack[v] = false;
    }
    public boolean hasCycle(){ return cycle != null; }
    public Iterable<Integer> cycle() { return cycle; }

    public static void main(String[] args) {
        // 输入用例参加附录1
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        // 输出结果参加附录6
        DirectedCycle directedCycle = new DirectedCycle(diGraph);
        System.out.println("有向图是否包含环: " + (directedCycle.hasCycle() ? "是" : "否"));
        if (directedCycle.hasCycle()){
            System.out.print("其中一条环为：");
            for (int i : directedCycle.cycle()) {
                System.out.print(i + " ");
            }
        }
        System.out.println();
    }
}

5 顶点的深度优先次序

顶点的深度优先次序分为前序、后序和逆后续，区别是记录点的时机发生在递归调用的前还是后。该算法产生的pre、post和reversePost

顺序在图的高级算法中十分有用。

public class DepthFirstOrder {
    private boolean[] marked;
    private ArrayDeque<Integer> pre;    // 保存前序遍历的结果
    private ArrayDeque<Integer> post;   // 保存后序的遍历结果
    private ArrayDeque<Integer> reversePost;    //保存逆后序的遍历结果
    public DepthFirstOrder(DiGraph G)
    {
        marked = new boolean[G.V()];
        pre = new ArrayDeque<>();
        post = new ArrayDeque<>();
        reversePost = new ArrayDeque<>();
        for (int v=0; v<G.V(); v++)
            if (!marked[v]) dfs(G, v);
    }
    private void dfs(DiGraph G, int v)
    {
        marked[v] = true;
        pre.add(v);
        for (int w : G.adj(v))
            if(!marked[w])
                dfs(G, w);
        post.add(v);
        // 按post的倒序保存
        reversePost.addFirst(v);
    }
    public Iterable<Integer> pre(){ return pre; }
    public Iterable<Integer> post(){ return post; }
    public Iterable<Integer> reversePost(){ return reversePost; }
    public static void main(String[] args) {
        // 构造无环图的输入参见附录7
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        DepthFirstOrder depthFirstOrder = new DepthFirstOrder(diGraph);
        // 输出结果参加附录8
        // 注意：对于同一幅图，构造图的输入顺序不一致
        // 会导致输出不相同
        System.out.print("前序节点顺序: ");
        for (int v : depthFirstOrder.pre())
            System.out.print(v + " ");
        System.out.println();
        System.out.print("后续节点顺序：");
        for (int v : depthFirstOrder.post())
            System.out.print(v + " ");
        System.out.println();
        System.out.print("逆后序节点顺序：");
        for (int v : depthFirstOrder.reversePost())
            System.out.print(v + " ");
    }
}

6 拓扑排序

给定一幅有向图，给出一组顶点排序，在有向图中，所有的边均是前面的点指向后面的点。

拓扑排序依赖图的环检测和逆后序遍历算法。

/**
 * 计算有向无环图中的所有顶点的拓扑排序，
 * 通常用于解决优先级限制下的调度问题
 */
public class Topological {
    private Iterable<Integer> order;
    public Topological(DiGraph G)
    {
        DirectedCycle directedCycle = new DirectedCycle(G);
        if(!directedCycle.hasCycle())
            order = new DepthFirstOrder(G).reversePost();
    }
    public boolean isDAG(){ return order == null; }
    public Iterable<Integer> order(){ return order; }
    public static void main(String[] args) {
        // 输入用例参考附录7
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        Topological topological = new Topological(diGraph);
        // 输出结果参见附录9
        if (topological.isDAG())
            System.out.println("有向图带有环，无法进行拓扑排序");
        else{
            System.out.print("拓扑排序结果：");
            for (int v : topological.order()) {
                System.out.print(v + " ");
            }
        }
    }
}

7 强联通检测

如果存在从v至w的路径，同时还存在从w至v的路径，则称v和w之间是强连通；如果一幅有向图中任意两点间都

是强连通，则这幅有向图也是强连通的。检测强连通算法依赖图的反转和逆后序遍历算法。算法比较简洁，但是

理解起来比较难，需要仔细分析理解。

/**
 * 有向图的强连通性，该算法依赖逆后序排序、图的反转、无向图的联通性算法
 */
public class SCC {
    private int[] id;
    private int count;
    private boolean[] marked;
    public SCC(DiGraph G)
    {
        id = new int[G.V()];
        marked = new boolean[G.V()];
        DepthFirstOrder depthFirstOrder = new DepthFirstOrder(G.reverse());
        for (int v : depthFirstOrder.reversePost())
            if(!marked[v]) {
                dfs(G, v);
                count++;
            }
    }
    private void dfs(DiGraph G, int v)
    {
        id[v] = count;
        marked[v] = true;
        for (int w : G.adj(v))
            if(!marked[w])
                dfs(G, w);
    }
    // 两点是否是强连通
    public boolean stronglyConnected(int v, int w){ return id[v] == id[w]; }
    // 强联通分量数
    public int count(){ return count; }
    // 节点所在的联通分量标识符
    public int id(int v){ return id[v]; }
    public static void main(String[] args) {
        // 带环的图，输入用例参见附录1
        DiGraph diGraph = new DiGraph(new Scanner(System.in));
        // 输出结果参见附录10
        SCC scc = new SCC(diGraph);
        System.out.println("有向图中强连通分量数：" + scc.count());
        System.out.println("节点6与12是否是强连通：" + (scc.stronglyConnected(6, 12) ? "是" : "否"));
        System.out.println("节点9与12是否是强连通：" + (scc.stronglyConnected(9, 12) ? "是" : "否"));
        System.out.println("输出联通分量");
        for (int i = 0; i < scc.count(); i++) {
            for (int v = 0; v < diGraph.V(); v++) {
                if(scc.id[v] == i)
                    System.out.print(v + " ");
            }
            System.out.println();
        }
    }
}

附录1，有向图构造数据

13
22
4 2
2 3
3 2
6 0
0 1
2 0
11 12
12 9
9 10
9 11
8 9
10 12
11 4
4 3
3 5
7 8
8 7
5 4
0 5
6 4
6 9
7 6

附录2，有向图输出

V: 13
E: 22
0: 1 5
1:
2: 3 0
3: 2 5
4: 2 3
5: 4
6: 0 4 9
7: 8 6
8: 9 7
9: 10 11
10: 12
11: 12 4
12: 9

附录3：有向图的可达性测试

顶点2到达的点
0 1 2 3 4 5
1，2，6能够到达的点
0 1 2 3 4 5 6 9 10 11 12

附录4：基于深度优先搜索的单向路径测试结果

顶点0可达的点: 0 1 2 3 4 5
0至12不存在有向路径
0至3的一条有向路径: 0 -> 5 -> 4 -> 2 -> 3

附录5：基于广度优先搜索的最短路径测试结果

顶点0可达的点: 0 1 2 3 4 5
0至12不存在有向路径
0至3的一条有向路径: 0 -> 5 -> 4 -> 3

附录6：检测环算法的测试输出

有向图是否包含环: 是
其中一条环为：3 2 4 5 3

附录7：构造无环图的输入用例

13
15
0 1
0 5
0 6
2 0
2 3
3 5
5 4
6 4
6 9
7 6
8 7
9 10
9 11
9 12
11 12

附录8：深度优先遍历图的输出结果

前序节点顺序: 0 1 5 4 6 9 10 11 12 2 3 7 8
后续节点顺序：1 4 5 10 12 11 9 6 0 3 2 7 8
逆后序节点顺序：8 7 2 3 0 6 9 11 12 10 5 4 1

附录9：拓扑排序测试输出结果

拓扑排序结果：8 7 2 3 0 6 9 11 12 10 5 4 1

附录10：带环有向图的强连通性测试输出结果

有向图中强连通分量数：5
节点6与12是否是强连通：否
节点9与12是否是强连通：是
输出联通分量
1
0 2 3 4 5
9 10 11 12
6
7 8