如何将强连通分量减少到一个顶点？

Question

来自https://algs4.cs.princeton.edu/42digraph/

43. 有向图中的可达顶点。设计一个线性时间算法来确定一个有向图是否有一个顶点可以从 每隔一个顶点。

Kosaraju-Sharir algorithm 为我们提供了强连接组件。可以看到here 的 Java 代码。将每个 SCC 缩减为一个顶点，一个出度为零的顶点可以相互到达。

问题是，似乎每个人都在谈论减少 SCC 而没有提供细节。这样做的有效算法是什么？

Answer 1

假设您已经有一种计算SCCs 的方法以及常用的图、顶点和边方法。然后它只是创建一个新图，为每个 SCC 添加一个顶点代表，然后添加边代表。

对于边，您需要能够将原始顶点（边目的地）映射到其在新图中的代表。您可以在第一遍中使用 Map<Vertex, SCC> 将顶点映射到它们的 SCC 和 Map<SCC, Vertex> 将 SCC 映射到新图中的代表顶点进行建模。或者你直接有一个Map<Vertex, Vertex> 将原始顶点映射到它们的代表。

---

这是一个 Java 解决方案：

public static Graph graphToSccGraph(Graph graph) {
    Collection<SCC> sccs = SccComputation.computeSccs(graph);
    Graph sccGraph = new Graph();

    Map<Vertex, SCC> vertexToScc = new HashMap<>();
    Map<SCC, Vertex> sccToRep = new HashMap<>();

    // Add a representative for each SCC (O(|V|))
    for (SCC scc : sccs) {
        Vertex rep = new Vertex();
        sccGraph.addVertex(rep);

        sccToRep.put(scc, rep);
        for (Vertex vertex : scc.getVertices()) {
            vertexToScc.put(vertex, scc);
        }
    }

    // Add edge representatives (O(|E|))
    for (Vertex vertex : graph.getVertices()) {
        Vertex sourceRep = sccToRep.get(vertexToScc.get(vertex));
        for (Edge edge : vertex.getOutgoingEdges()) {
           Vertex destRep = sccToRep.get(vertexToScc.get(edge.getDestination()));
           Edge edgeRep = new Edge(sourceRep, destRep);
              if (!sccGraph.contains(edgeRep)) {
                  sccGraph.addEdge(edgeRep);
              }
        }
    }

    return sccGraph;
}

---

时间复杂度与图的大小（顶点和边的数量）呈线性，因此最优。那是Theta(|V| + |E|)。

通常人们使用Union-Find（参见Wikipedia）数据结构来简化此操作并摆脱Maps。

Answer 2

以下是我自己的问题的 Java 解决方案。对于图形表示，它使用来自https://github.com/kevin-wayne/algs4 的edu.princeton.cs:algs4:1.0.3。如this paper 所述，似乎有用于图形收缩的通用算法；但是，就我的目的而言，以下内容就足够了。

/**
 * 43. <b>Reachable vertex.</b>
 * <p>
 * DAG: Design a linear-time algorithm to determine whether a DAG has a vertex that is reachable from every other
 * vertex, and if so, find one.
 * Digraph: Design a linear-time algorithm to determine whether a digraph has a vertex that is reachable from every
 * other vertex, and if so, find one.
 * <p>
 * Answer:
 * DAG: Consider an edge (u, v) ∈ E. Since the graph is acyclic, u is not reachable from v.
 * Thus u cannot be the solution to the problem. From this it follows that only a vertex of
 * outdegree zero can be a solution. Furthermore, there has to be exactly one vertex with outdegree zero,
 * or the problem has no solution. This is because if there were multiple vertices with outdegree zero,
 * they wouldn't be reachable from each other.
 * <p>
 * Digraph: Reduce the graph to it's Kernel DAG, then find a vertex of outdegree zero.
 */
public class Scc {
    private final Digraph g;
    private final Stack<Integer> s = new Stack<>();
    private final boolean marked[];
    private final Digraph r;
    private final int[] scc;
    private final Digraph kernelDag;

    public Scc(Digraph g) {
        this.g = g;
        this.r = g.reverse();
        marked = new boolean[g.V()];
        scc = new int[g.V()];
        Arrays.fill(scc, -1);

        for (int v = 0; v < r.V(); v++) {
            if (!marked[v]) visit(v);
        }

        int i = 0;
        while (!s.isEmpty()) {
            int v = s.pop();

            if (scc[v] == -1) visit(v, i++);
        }
        Set<Integer> vPrime = new HashSet<>();
        Set<Map.Entry<Integer, Integer>> ePrime = new HashSet<>();

        for (int v = 0; v < scc.length; v++) {
            vPrime.add(scc[v]);
            for (int w : g.adj(v)) {
                // no self-loops, no parallel edges
                if (scc[v] != scc[w]) {
                    ePrime.add(new SimpleImmutableEntry<>(scc[v], scc[w]));
                }
            }
        }
        kernelDag = new Digraph(vPrime.size());
        for (Map.Entry<Integer, Integer> e : ePrime) kernelDag.addEdge(e.getKey(), e.getValue());
    }

    public int reachableFromAllOther() {
        for (int v = 0; v < kernelDag.V(); v++) {
            if (kernelDag.outdegree(v) == 0) return v;
        }
        return -1;
    }

    // reverse postorder
    private void visit(int v) {
        marked[v] = true;

        for (int w : r.adj(v)) {
            if (!marked[w]) visit(w);
        }
        s.push(v);
    }

    private void visit(int v, int i) {
        scc[v] = i;

        for (int w : g.adj(v)) {
            if (scc[w] == -1) visit(w, i);
        }
    }
}

在下图中运行它会产生如图所示的强连接组件。简化 DAG 中的顶点 0 可以从其他每个顶点到达。

我在任何地方都找不到我上面介绍的那种细节。诸如“嗯，这很容易，你这样做，然后你再做其他事情”之类的评论在没有具体细节的情况下被抛出。