有向图

有向图的数据结构

采用链表

public class Digraph {
    private static final String NEWLINE = System.getProperty("line.separator");
    private final int V;           // number of vertices in this digraph
    private int E;                 // number of edges in this digraph
    private Bag<Integer>[] adj;    // adj[v] = adjacency list for vertex v
    private int[] indegree;        // indegree[v] = indegree of vertex v
    public Digraph(int V) {
        if (V < 0) throw new IllegalArgumentException("Number of vertices in a Digraph must be nonnegative");
        this.V = V;
        this.E = 0;
        indegree = new int[V];
        adj = (Bag<Integer>[]) new Bag[V];
        for (int v = 0; v < V; v++) {
            adj[v] = new Bag<Integer>();
        }
    }
    public Digraph(In in) {
        try {
            this.V = in.readInt();
            if (V < 0) throw new IllegalArgumentException("Number of vertices in a Digraph must be nonnegative");
            indegree = new int[V];
            adj = (Bag<Integer>[]) new Bag[V];
            for (int v = 0; v < V; v++) {
                adj[v] = new Bag<Integer>();
            }
            int E = in.readInt();
            if (E < 0) throw new IllegalArgumentException("Number of edges in a Digraph must be nonnegative");
            for (int i = 0; i < E; i++) {
                int v = in.readInt();
                int w = in.readInt();
                addEdge(v, w);
            }
        }
        catch (NoSuchElementException e) {
            throw new InputMismatchException("Invalid input format in Digraph constructor");
        }
    }
    public Digraph(Digraph G) {
        this(G.V());
        this.E = G.E();
        for (int v = 0; v < V; v++)
            this.indegree[v] = G.indegree(v);
        for (int v = 0; v < G.V(); v++) {
            // reverse so that adjacency list is in same order as original
            Stack<Integer> reverse = new Stack<Integer>();
            for (int w : G.adj[v]) {
                reverse.push(w);
            }
            for (int w : reverse) {
                adj[v].add(w);
            }
        }
    }
    public int V() {
        return V;
    }
    public int E() {
        return E;
    }
    // throw an IndexOutOfBoundsException unless 0 <= v < V
    private void validateVertex(int v) {
        if (v < 0 || v >= V)
            throw new IndexOutOfBoundsException("vertex " + v + " is not between 0 and " + (V-1));
    }
    public void addEdge(int v, int w) {
        validateVertex(v);
        validateVertex(w);
        adj[v].add(w);
        indegree[w]++;
        E++;
    }
    public Iterable<Integer> adj(int v) {
        validateVertex(v);
        return adj[v];
    }
    public int outdegree(int v) {
        validateVertex(v);
        return adj[v].size();
    }
    public int indegree(int v) {
        validateVertex(v);
        return indegree[v];
    }
    public Digraph reverse() {
        Digraph R = new Digraph(V);
        for (int v = 0; v < V; v++) {
            for (int w : adj(v)) {
                R.addEdge(w, v);
            }
        }
        return R;
    }
    public String toString() {
        StringBuilder s = new StringBuilder();
        s.append(V + " vertices, " + E + " edges " + NEWLINE);
        for (int v = 0; v < V; v++) {
            s.append(String.format("%d: ", v));
            for (int w : adj[v]) {
                s.append(String.format("%d ", w));
            }
            s.append(NEWLINE);
        }
        return s.toString();
    }
}

有向图的可达性（采用深度优先搜索）

两种构造函数分别针对单源可达性和多源可达性

public class DirectedDFS {
    private boolean[] marked;  // marked[v] = true if v is reachable
                               // from source (or sources)
    private int count;         // number of vertices reachable from s
    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) {
        count++;
        marked[v] = true;
        for (int w : G.adj(v)) {
            if (!marked[w]) dfs(G, w);
        }
    }
    public boolean marked(int v) {
        return marked[v];
    }
    public int count() {
        return count;
    }
}

有向图中检测环

public class DirectedCycle {
    private boolean[] marked;        // marked[v] = has vertex v been marked?
    private int[] edgeTo;            // edgeTo[v] = previous vertex on path to v
    private boolean[] onStack;       // onStack[v] = is vertex on the stack?
    private Stack<Integer> cycle;    // directed cycle (or null if no such cycle)
    public DirectedCycle(Digraph G) {
        marked  = new boolean[G.V()];
        onStack = new boolean[G.V()];
        edgeTo  = new int[G.V()];
        for (int v = 0; v < G.V(); v++)
            if (!marked[v] && cycle == null) dfs(G, v);
    }
    // check that algorithm computes either the topological order or finds a directed cycle
    private void dfs(Digraph G, int v) {
        onStack[v] = true;
        marked[v] = true;
        for (int w : G.adj(v)) {

            // short circuit if directed cycle found
            if (cycle != null) return;

            //found new vertex, so recur
            else if (!marked[w]) {
                edgeTo[w] = v;
                dfs(G, w);
            }

            // trace back directed cycle
            else if (onStack[w]) {
                cycle = new Stack<Integer>();
                for (int x = v; x != w; x = edgeTo[x]) {
                    cycle.push(x);
                }
                cycle.push(w);
                cycle.push(v);
                assert check();
            }
        }
        onStack[v] = false;
    }
    public boolean hasCycle() {
        return cycle != null;
    }
    public Iterable<Integer> cycle() {
        return cycle;
    }
    // certify that digraph has a directed cycle if it reports one
    private boolean check() {

        if (hasCycle()) {
            // verify cycle
            int first = -1, last = -1;
            for (int v : cycle()) {
                if (first == -1) first = v;
                last = v;
            }
            if (first != last) {
                System.err.printf("cycle begins with %d and ends with %d\n", first, last);
                return false;
            }
        }

        return true;
    }
}

拓扑排序

给定一个有向图，将所有的顶点排序，使得所有的有向边均是从前面的元素指向排在后面的元素（序列前后某两点并不一定是连通的，但是连通的每条边都是从前面某点指向这个序列的后面某点）。

当且仅当一幅有向图是无环图时它才能进行拓扑排序（有环就分不清前后了，还排什么序）

基于深度优先搜索的顶点排序（不是拓扑排序，十分重要）

先学习顶点排序：这里面有preourder，postorder和reverse postorder

preorder也叫前序，是dfs调用的顺序（每次调用先给你加入队列，追踪dfs的调用顺序）

postorder也叫后序，是顶点遍历完成后（先走到尽头或者说先完成dfs(那个for结束了)）的顺序

reverse postorder也叫逆后序，就是后序倒过来

public class DepthFirstOrder {
    private boolean[] marked;          // marked[v] = has v been marked in dfs?
    private int[] pre;                 // pre[v]    = preorder  number of v
    private int[] post;                // post[v]   = postorder number of v
    private Queue<Integer> preorder;   // vertices in preorder
    private Queue<Integer> postorder;  // vertices in postorder
    private int preCounter;            // counter or preorder numbering
    private int postCounter;           // counter for postorder numbering
    public DepthFirstOrder(Digraph G) {
        pre    = new int[G.V()];
        post   = new int[G.V()];
        postorder = new Queue<Integer>();
        preorder  = new Queue<Integer>();
        marked    = new boolean[G.V()];
        for (int v = 0; v < G.V(); v++)
            if (!marked[v]) dfs(G, v);
    }
    public DepthFirstOrder(EdgeWeightedDigraph G) {
        pre    = new int[G.V()];
        post   = new int[G.V()];
        postorder = new Queue<Integer>();
        preorder  = new Queue<Integer>();
        marked    = new boolean[G.V()];
        for (int v = 0; v < G.V(); v++)
            if (!marked[v]) dfs(G, v);
    }

    // run DFS in digraph G from vertex v and compute preorder/postorder
    private void dfs(Digraph G, int v) {
        marked[v] = true;
        pre[v] = preCounter++;
        preorder.enqueue(v);
        for (int w : G.adj(v)) {
            if (!marked[w]) {
                dfs(G, w);
            }
        }
        postorder.enqueue(v);
        post[v] = postCounter++;
    }
    // run DFS in edge-weighted digraph G from vertex v and compute preorder/postorder
    private void dfs(EdgeWeightedDigraph G, int v) {
        marked[v] = true;
        pre[v] = preCounter++;
        preorder.enqueue(v);
        for (DirectedEdge e : G.adj(v)) {
            int w = e.to();
            if (!marked[w]) {
                dfs(G, w);
            }
        }
        postorder.enqueue(v);
        post[v] = postCounter++;
    }
    public int pre(int v) {
        return pre[v];
    }
    public int post(int v) {
        return post[v];
    }
    public Iterable<Integer> post() {
        return postorder;
    }
    public Iterable<Integer> pre() {
        return preorder;
    }
    public Iterable<Integer> reversePost() {
        Stack<Integer> reverse = new Stack<Integer>();
        for (int v : postorder)
            reverse.push(v);
        return reverse;
    }
    // check that pre() and post() are consistent with pre(v) and post(v)
    private boolean check(Digraph G) {
        // check that post(v) is consistent with post()
        int r = 0;
        for (int v : post()) {
            if (post(v) != r) {
                StdOut.println("post(v) and post() inconsistent");
                return false;
            }
            r++;
        }
        // check that pre(v) is consistent with pre()
        r = 0;
        for (int v : pre()) {
            if (pre(v) != r) {
                StdOut.println("pre(v) and pre() inconsistent");
                return false;
            }
            r++;
        }
        return true;
    }
}

拓扑排序

顶点排序的逆后序就是拓扑排序（因为顶点后序中的点是先完成dfs（或者叫先走到尽头）的点排在前面，在dfs每到尽头回溯的过程中，排在前面的点（走到尽头）肯定能被排在后面的（往回 回溯的）某些点访问到，所以倒过来，这个肯定能被访问到的点就放在拓扑排序的靠后面了，相当于拓扑排序把后序倒了过来）

内部就是使用了DepthFirstOrder

public class Topological {
    private Iterable<Integer> order;  // topological order
    private int[] rank;               // rank[v] = position of vertex v in topological order
    public Topological(Digraph G) {
        DirectedCycle finder = new DirectedCycle(G);
        if (!finder.hasCycle()) {
            DepthFirstOrder dfs = new DepthFirstOrder(G);
            order = dfs.reversePost();
            rank = new int[G.V()];
            int i = 0;
            for (int v : order)
                rank[v] = i++;
        }
    }
    public Topological(EdgeWeightedDigraph G) {
        EdgeWeightedDirectedCycle finder = new EdgeWeightedDirectedCycle(G);
        if (!finder.hasCycle()) {
            DepthFirstOrder dfs = new DepthFirstOrder(G);
            order = dfs.reversePost();
        }
    }
    public Iterable<Integer> order() {
        return order;
    }
    public boolean hasOrder() {
        return order != null;
    }
    public int rank(int v) {
        validateVertex(v);
        if (hasOrder()) return rank[v];
        else            return -1;
    }
    // throw an IndexOutOfBoundsException unless 0 <= v < V
    private void validateVertex(int v) {
        int V = rank.length;
        if (v < 0 || v >= V)
            throw new IndexOutOfBoundsException("vertex " + v + " is not between 0 and " + (V-1));
    }
}

有向图中的强连通性

强连通： 如果两个顶点v和w是互相可达的，则称它们为强连通的

两个顶点是强连通的当且仅当它们都在一个普通的有向环中

强连通的性质（强连通是一种等价关系）：

自反性：任意顶点到自己都是强连通的
对称性
传递性

强连通分量： 强连通将所有顶点分为了一些等价的部分，每个部分都由互相均为强连通的顶点的最大子集组成的。我们将这些子集称为强连通分量。例如下图含有5个强连通分量：

强连通分量是基于顶点的，不是基于边的，有些边的两个顶点可能在不同的强连通分量中

Kosaraju算法

代码很简单:按照 反向图顶点排序的 逆后序的顺序再做dfs就行了（虽然一次dfs只能证明s->v,但是由于采用反向图的逆后序，v->s已经天然证明了（而且这个v->s的证明还用了前面的s->v的结果！））

public class KosarajuSharirSCC {
    private boolean[] marked;     // marked[v] = has vertex v been visited?
    private int[] id;             // id[v] = id of strong component containing v
    private int count;            // number of strongly-connected components
    public KosarajuSharirSCC(Digraph G) {
        // compute reverse postorder of reverse graph
        DepthFirstOrder dfs = new DepthFirstOrder(G.reverse());
        // run DFS on G, using reverse postorder to guide calculation
        marked = new boolean[G.V()];
        id = new int[G.V()];
        for (int v : dfs.reversePost()) {
            if (!marked[v]) {
                dfs(G, v);
                count++;
            }
        }
        // check that id[] gives strong components
        assert check(G);
    }
    // DFS on graph G
    private void dfs(Digraph G, int v) {
        marked[v] = true;
        id[v] = count;
        for (int w : G.adj(v)) {
            if (!marked[w]) dfs(G, w);
        }
    }
    public int count() {
        return count;
    }
    public boolean stronglyConnected(int v, int w) {
        return id[v] == id[w];
    }
    public int id(int v) {
        return id[v];
    }
    // does the id[] array contain the strongly connected components?
    private boolean check(Digraph G) {
        TransitiveClosure tc = new TransitiveClosure(G);
        for (int v = 0; v < G.V(); v++) {
            for (int w = 0; w < G.V(); w++) {
                if (stronglyConnected(v, w) != (tc.reachable(v, w) && tc.reachable(w, v)))
                    return false;
            }
        }
        return true;
    }
}

传递闭包

上面的check方法有一个传递闭包类，就是判断两点的是否连通（能否到达）

传递闭包就是另一种图（这个图(存图的二维矩阵中)的每个位置不是存直接相连的边，而是存 “两者是否相连”的一个布尔值，能连通就是true）

类中构造函数所需的空间和V2成正比，时间和V(V+E)成正比

public class TransitiveClosure {
    private DirectedDFS[] tc;  // tc[v] = reachable from v
    public TransitiveClosure(Digraph G) {
        tc = new DirectedDFS[G.V()];
        for (int v = 0; v < G.V(); v++)
            tc[v] = new DirectedDFS(G, v);
    }
    public boolean reachable(int v, int w) {
        return tc[v].marked(w);
    }
}