有向图的基本算法

　　有向图的基本定义：由一组顶点和一组有向边组成，每条有向边连接着有序的一对顶点。

import java.util.InputMismatchException;
import java.util.NoSuchElementException;

public class Digraph {
    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 int V() {
        return V;
    }

    public int E() {
        return E;
    }

    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 reverse = new Digraph(V);
        for (int v = 0; v < V; v++) {
            for (int w : adj(v)) {
                reverse.addEdge(w, v);
            }
        }
        return reverse;
    } 
}

　　有向路径由一系列顶点组成，对于其中的每个顶点都存在有向边从它指向序列中的下个顶点。

　　有向环：一条至少含有一条边且起点和重点相同的有向路径。　　

有向图的可达性　　

　　单点可达性问题给定一幅有向图和一个起点s，回答是否存在一条从s到给定顶点v的有向路径。

　　多点可达性给定一幅有向图图和顶点的集合，回答是否存在一条从集合中的任意顶点到达给定顶点V的有向路径。

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;
    }
}

　　有向无环图（DAG）：就是一幅不含有环的有向图。　

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);
    }

    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;
    }
}

　　有向图的拓扑排序： 对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序，是将G中所有顶点排成一个线性序列，使得图中任意一对顶点u和v，若<u，v> ∈E(G)，则u在线性序列中出现在v之前。 通常，这样的线性序列称为满足拓扑次序(Topological Order)的序列，简称拓扑序列。

public class Topological {
    private Iterable<Integer> order;  // topological order
    private int[] rank;       // rank[v] = position of vertex v in topological order

    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) {
        if (hasOrder()) return rank[v];
        else            return -1;
    }

}

　　有向图中基于深度优先搜索的顶点排序:

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(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);
    }

    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;
    } 
}

　　顶点可达性问题：是否存在一条从给定顶点V到另一个给定顶点W的路径。

　　强连通性：如果两个顶点V和W是互相可达的，则称它们为强连通。强连通具有以下性质：

1. 自反性：任意顶点V和自己都是强连通的。
2. 对称性：如果V和W是强连通的，那么W和V也是强连通的。
3. 传递性：如果V和W是强连通的且W和X是强连通的，那么V和X是强连通的。

　　强连通分量：每个部分都是互为强连通的顶点的最大子集组成。

Kosaraju算法

/******************************************************************************
 *  Compilation:  javac KosarajuSharirSCC.java
 *  Execution:    java KosarajuSharirSCC filename.txt
 *  Dependencies: Digraph.java TransitiveClosure.java StdOut.java In.java
 *  Data files:   http://algs4.cs.princeton.edu/42digraph/tinyDG.txt
 *
 *  Compute the strongly-connected components of a digraph using the
 *  Kosaraju-Sharir algorithm.
 *
 *  Runs in O(E + V) time.
 *
 *  % java KosarajuSCC tinyDG.txt
 *  5 components
 *  1 
 *  0 2 3 4 5 
 *  9 10 11 12 
 *  6 8 
 *  7
 *
 *  % java KosarajuSharirSCC mediumDG.txt 
 *  10 components
 *  21 
 *  2 5 6 8 9 11 12 13 15 16 18 19 22 23 25 26 28 29 30 31 32 33 34 35 37 38 39 40 42 43 44 46 47 48 49 
 *  14 
 *  3 4 17 20 24 27 36 
 *  41 
 *  7 
 *  45 
 *  1 
 *  0 
 *  10 
 *
 *  % java -Xss50m KosarajuSharirSCC mediumDG.txt 
 *  25 components
 *  7 11 32 36 61 84 95 116 121 128 230   ...
 *  28 73 80 104 115 143 149 164 184 185  ...
 *  38 40 200 201 207 218 286 387 418 422 ...
 *  12 14 56 78 87 103 216 269 271 272    ...
 *  42 48 112 135 160 217 243 246 273 346 ...
 *  46 76 96 97 224 237 297 303 308 309   ...
 *  9 15 21 22 27 90 167 214 220 225 227  ...
 *  74 99 133 146 161 166 202 205 245 262 ...
 *  43 83 94 120 125 183 195 206 244 254  ...
 *  1 13 54 91 92 93 106 140 156 194 208  ...
 *  10 39 67 69 131 144 145 154 168 258   ...
 *  6 52 66 113 118 122 139 147 212 213   ...
 *  8 127 150 182 203 204 249 367 400 432 ...
 *  63 65 101 107 108 136 169 170 171 173 ...
 *  55 71 102 155 159 198 228 252 325 419 ...
 *  4 25 34 58 70 152 172 196 199 210 226 ...
 *  2 44 50 88 109 138 141 178 197 211    ...
 *  57 89 129 162 174 179 188 209 238 276 ...
 *  33 41 49 119 126 132 148 181 215 221  ...
 *  3 18 23 26 35 64 105 124 157 186 251  ...
 *  5 16 17 20 31 47 81 98 158 180 187    ...
 *  24 29 51 59 75 82 100 114 117 134 151 ...
 *  30 45 53 60 72 85 111 130 137 142 163 ...
 *  19 37 62 77 79 110 153 352 353 361    ...
 *  0 68 86 123 165 176 193 239 289 336   ...
 *
 ******************************************************************************/

/**
 *  The <tt>KosarajuSharirSCC</tt> class represents a data type for 
 *  determining the strong components in a digraph.
 *  The <em>id</em> operation determines in which strong component
 *  a given vertex lies; the <em>areStronglyConnected</em> operation
 *  determines whether two vertices are in the same strong component;
 *  and the <em>count</em> operation determines the number of strong
 *  components.

 *  The <em>component identifier</em> of a component is one of the
 *  vertices in the strong component: two vertices have the same component
 *  identifier if and only if they are in the same strong component.

 *  <p>
 *  This implementation uses the Kosaraju-Sharir algorithm.
 *  The constructor takes time proportional to <em>V</em> + <em>E</em>
 *  (in the worst case),
 *  where <em>V</em> is the number of vertices and <em>E</em> is the number of edges.
 *  Afterwards, the <em>id</em>, <em>count</em>, and <em>areStronglyConnected</em>
 *  operations take constant time.
 *  For alternate implementations of the same API, see
 *  {@link TarjanSCC} and {@link GabowSCC}.
 *  <p>
 *  For additional documentation,
 *  see <a href="http://algs4.cs.princeton.edu/42digraph">Section 4.2</a> of
 *  <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
 *
 *  @author Robert Sedgewick
 *  @author Kevin Wayne
 */
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

    /**
     * Computes the strong components of the digraph <tt>G</tt>.
     * @param G the digraph
     */
    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);
        }
    }

    /**
     * Returns the number of strong components.
     * @return the number of strong components
     */
    public int count() {
        return count;
    }

    /**
     * Are vertices <tt>v</tt> and <tt>w</tt> in the same strong component?
     * @param v one vertex
     * @param w the other vertex
     * @return <tt>true</tt> if vertices <tt>v</tt> and <tt>w</tt> are in the same
     *     strong component, and <tt>false</tt> otherwise
     */
    public boolean stronglyConnected(int v, int w) {
        return id[v] == id[w];
    }

    /**
     * Returns the component id of the strong component containing vertex <tt>v</tt>.
     * @param v the vertex
     * @return the component id of the strong component containing vertex <tt>v</tt>
     */
    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;
    }

    /**
     * Unit tests the <tt>KosarajuSharirSCC</tt> data type.
     */
    public static void main(String[] args) {
        In in = new In(args[0]);
        Digraph G = new Digraph(in);
        KosarajuSharirSCC scc = new KosarajuSharirSCC(G);

        // number of connected components
        int m = scc.count();
        StdOut.println(m + " components");

        // compute list of vertices in each strong component
        Queue<Integer>[] components = (Queue<Integer>[]) new Queue[m];
        for (int i = 0; i < m; i++) {
            components[i] = new Queue<Integer>();
        }
        for (int v = 0; v < G.V(); v++) {
            components[scc.id(v)].enqueue(v);
        }

        // print results
        for (int i = 0; i < m; i++) {
            for (int v : components[i]) {
                StdOut.print(v + " ");
            }
            StdOut.println();
        }

    }

}