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Algorithm-and-Leetcode/DataStructrue/Graph/DirectedGraph.md at master · Seanforfun/Algorithm-and-Leetcode

Directed Graph

有向图，边是单向的：每条边所连接的两个顶点都是一个有序对，它的邻接性是单向的。 我们称一条有向边由第一个顶点指出并指向第二个顶点。 一个顶点的出度(out degree)由该顶点指出的边的总数。 一个顶点的入度(in degree)为指向该顶点的边的总数。 一条有向边的第一个顶点称为它的头，第二个顶点称为它的尾。

接口

public interface Digraph {
	/**
	 * @Description: Number of vertex.
	 * @return
	 */
	public int V();
	/**
	 * @Description: Number of edge.
	 * @return
	 */
	public int E();
	/**
	 * @Description: Add an edge between two vertex. v->w.
	 * @param v
	 * @param w
	 */
	public void addEdge(int v, int w);
	/**
	 * @Description: Return vertex point out from v.
	 * @param v
	 * @return
	 */
	public Iterable<Integer> adj(int v);
	/**
	 * @Description: Get the reverse graph of current graph.
	 * @return
	 */
	public Digraph reverse();
	/**
	 * @Description: Print current graph.
	 */
	public void display();
}

实现类

public class DigraphImpl implements Digraph {
	private final int V;		//Number of vertex in current digraph
	private int E; 	//Number of edges.
	private ListBag<Integer>[] adj;
	@SuppressWarnings("unchecked")
	public DigraphImpl(int V){
		this.V = V;
		adj = (ListBag<Integer>[])new ListBag[V];
		for (int i = 0; i < V; i++) {
			adj[i] = new ListBag<Integer>(); 
		}
		this.E = 0;
	}
	
	@SuppressWarnings("unchecked")

	public DigraphImpl(FileInputStream in) {
		Scanner s = null;
		s = new Scanner(in);
		this.V = s.nextInt();
		this.E = s.nextInt();
		adj = new ListBag[V];
		for (int i = 0; i < V; i++) {
			adj[i] = new ListBag<Integer>(); 
		}
		for(int i = 0; i < E; i ++){
			int v = s.nextInt();
			int w = s.nextInt();
			adj[v].add(w);
		}
		s.close();
	}
	@Override
	public int V() { return this.V; }
	@Override
	public void addEdge(int v, int w) {	//只增加一条有向边
		adj[v].add(w);
		E++;
	}
	@Override
	public int E() { return E; }
	@Override
	public Iterable<Integer> adj(int v) {
		return adj[v];
	}
	@Override
	public Digraph reverse() {
		DigraphImpl g  = new DigraphImpl(V);
		for(int v = 0; v < V; v++)
			for(int w : adj(v))
				addEdge(w, v);
		return g;
	}
	@Override
	public void display() {
		for(int i = 0; i < V; i++){
			StringBuilder sb = new StringBuilder(i + ": ");
			for(int w : adj[i]){
				sb.append(w + "|");
			}
			System.out.println(sb.toString());
		}
	}
}

测试

	public static void main(String[] args) throws IOException {
		FileInputStream in = new FileInputStream(new File("src/ca/mcmaster/chapter/four/graph/tinyDG.txt"));
		Digraph graph = new DigraphImpl(in);
		in.close();
		graph.display();
	}

结果

0: 5|1| 1: 2: 0|3| 3: 5|2| 4: 3|2| 5: 4| 6: 9|4|8|0| 7: 6|9| 8: 6| 9: 11|10| 10: 12| 11: 4|12| 12: 9|

有向图的可达性

可达性的重要实际应用是在内存管理系统中，一个顶点表示一个对象，一条边表示一个对象的引用。

public class DirectedDFS {
	private final boolean[] marked;
	public DirectedDFS(Digraph g, int s){
		this.marked = new boolean[g.V()];
		dfs(g, s);
	}
	public DirectedDFS(Digraph g, Iterable<Integer> sources){
		marked = new boolean[g.V()];
		for(int s : sources)
			if(!marked[s]) dfs(g, s);
	}
	private void dfs(Digraph g, int s) {
		marked[s] = true;
		for(int w : g.adj(s))
			if(!marked[w])	dfs(g, w);
	}
	public boolean mark(int v){
		return marked[v];
	}
}
  • 测试
public static void main(String[] args) throws FileNotFoundException {
		Digraph g = new DigraphImpl(new FileInputStream(new File("src/ca/mcmaster/chapter/four/graph/tinyDG.txt")));
		Bag<Integer> bag = new ListBag<Integer>();
		bag.add(1);
		bag.add(2);
		bag.add(6);
		DirectedDFS d = new DirectedDFS(g, bag);
		StringBuilder sb = new StringBuilder();
		for(int v = 0; v < g.V(); v++){
			if(d.mark(v)) sb.append(v + " ");
		}
		System.out.println(sb.toString());
	}

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

单点有向路径

通过DFPath实现

public class DepthFirstPathDirectedGraph implements Path {
	private final DigraphImpl g;
	private int s;
	private final boolean[] marked;	//Used to mark if current vertex has been accessed.
	private final int[] edgeTo;		//Used to save the vertex ahead of current vertex.
	public DepthFirstPathDirectedGraph(DigraphImpl g, int s){
		this.s = s;
		this.g = (DigraphImpl) g;
		marked = new boolean[g.V()];
		edgeTo = new int[g.V()];
		dfs(g, s);
	}
	public DepthFirstPathDirectedGraph(String file, int s) throws FileNotFoundException{
		g = new DigraphImpl(new FileInputStream(new File(file)));
		this.s = s;
		marked = new boolean[g.V()];
		edgeTo = new int[g.V()];
		dfs(g, s);
	}
	@Override
	public boolean hasPathTo(int v) {
		return edgeTo[v] != 0;
	}
	@Override
	public Iterable<Integer> pathTo(int v) {
		Bag<Integer> path = new ListBag<Integer>();
		path.add(v);
		while(edgeTo[v] != s){
			path.add(edgeTo[v]);
			v = edgeTo[v];
		}
		return path;
	}
	private void dfs(DigraphImpl g, int v){
		marked[v] = true;
		for(int w : g.adj(v)){
			if(!marked[w]){
				edgeTo[w] = v;
				dfs(g, w);
			}
		}
	}
}
  • 测试
	public static void main(String[] args) throws FileNotFoundException {
		DigraphImpl g = new DigraphImpl(new FileInputStream(new File("src/ca/mcmaster/chapter/four/graph/tinyDG.txt")));
		DepthFirstPathDirectedGraph p = new DepthFirstPathDirectedGraph(g, 6);
		Iterable<Integer> pathTo = p.pathTo(0);
		System.out.print(6 + " ");
		for(Integer w : pathTo)	System.out.print(w + " ");
		System.out.println();
//		System.out.println(p.hasPathTo(10));
	}

6 9 11 4 3 2 0

通过BFPath实现

public class BreadFirstPathDirectedGraph implements Path {
	private final Digraph g;
	private int s;
	private final boolean[] marked;
	private final int[] edgeTo;
	public BreadFirstPathDirectedGraph(Digraph g, int s) {
		this.g = g;
		this.s = s;
		marked = new boolean[g.V()];
		edgeTo = new int[g.V()];
		bfs(g, s);
	}
	@Override
	public boolean hasPathTo(int v) {
		return marked[v];
	}
	@Override
	public Iterable<Integer> pathTo(int v) {
		Stack<Integer> path = new Stack<>();
		path.push(v);
		while(s != edgeTo[v]){
			path.push(edgeTo[v]);
			v= edgeTo[v];
		}
		return path;
	}
	private void bfs(Digraph g, int s){
		LinkedBlockingQueue<Integer> q = new LinkedBlockingQueue<>();
		marked[s] = true;
		q.offer(s);
		while(!q.isEmpty()){
			int v = q.poll();
			for(int w : g.adj(v)){
				if(!marked[w]){
					edgeTo[w] = v;
					marked[w] = true;
					q.offer(w);
				}
			}
		}
	}
}
  • 测试和DFS完全一致。

寻找有向环

public class DirectedCycle {
	private final boolean[] marked;
	private final int[] edgeTo;
	private Stack<Integer> cycle = null;	//有向环上的所有顶点
	private final boolean[] onStack;	//递归调用的栈上的所有顶点
	public DirectedCycle(Digraph g){
		onStack = new boolean[g.V()];
		marked = new boolean[g.V()];
		edgeTo = new int[g.V()];
		int size = g.V();
		for(int v = 0; v < size; v++)
			if(!marked[v])	dfs(g, v);
	}
	private void dfs(Digraph g, int v){
		//在此次的遍历中，记录当前的位置可能在环上。如果结束了没有在环上则改回false
		onStack[v] = true;
		marked[v] = true;
		for(int w : g.adj(v)){
			if(this.hasCycle()) return;
			else if(!marked[w]){	//当前结点未被访问过，一定不是闭环，递归调用
				edgeTo[w] = v;
				dfs(g, w);
			}
			else if(onStack[w]){		//在此次的遍历中，当前点已经在环上，说明已经构成了cycle.v的下一个结点构成了闭环。
				//创建stack对象，并添加元素。
				cycle = new Stack<>();
				for(int x = v; x != w; x = edgeTo[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;
	}
}
  • 测试
	public static void main(String[] args) throws FileNotFoundException {
		Digraph g = new DigraphImpl(new FileInputStream(new File("src/ca/mcmaster/chapter/four/graph/tinyDG.txt")));
		DirectedCycle cycle = new DirectedCycle(g);
		StringBuilder sb = new StringBuilder();
		for(int i : cycle.cycle()){
			sb.append(i + " ");
		}
		System.out.println(sb.toString());
	}

3 4 5 3

有向图的遍历

  • 有向图的遍历分成三种。
  1. 前序遍历

前序遍历就是dfs()的调用顺序，将顶点加入队列。

  1. 后序遍历

后序遍历，就是递归调用之后将顶点加入队列中。

  1. 逆后序遍历

逆后序遍历和后序遍历相似，但是是将顶点加入栈中。

public class DepthFirstOrder {
	private final boolean[] marked;
	private final Queue<Integer> pre;	//前序遍历。和dfs的顺序一致，当访问到了一个顶点，就将顶点加入队列。
	private final Queue<Integer> post;	//后序遍历，顶点遍历完成的顺序。从递归的最内部开始加入遍历，相当于完成后立刻加入队列之中。
	private final Stack<Integer> reversePost;
	public DepthFirstOrder(Digraph g){
		pre = new LinkedBlockingQueue<Integer>();
		post = new LinkedBlockingQueue<>();
		reversePost = new Stack<>();
		marked = new boolean[g.V()];
		for(int v = 0; v < g.V(); v++)
			if(!marked[v]) dfs(g, v);
	}
	private void dfs(Digraph g, int v){
		pre.offer(v);	//前序：一访问到某个顶点就立刻加入队列。
		marked[v] = true;
		for(int w : g.adj(v))
			if(!marked[w])	dfs(g, w);
		post.offer(v);	//后序：完成后，将顶点加入队列。从递归的最内部加入队列。
		reversePost.push(v);	//逆后序：加入栈
	}
	public Iterable<Integer> pre(){
		return this.pre;
	}
	public Iterable<Integer> post(){
		return this.post();
	}
	public Iterable<Integer> reversePost(){
		return this.reversePost;
	}
}

拓扑排序

给定一张有向图，将所有的顶点排序，使得所有的有向边均从排在前面的元素指向排在后面的元素。 有环图画不出拓扑图，因为有环图无法确定环上定点的顺序。 一幅有向无环图的拓扑顺序即为所有顶点的逆后序排列。

public class Topological {
	private Iterable<Integer> order;
	public Topological(Digraph g){
		DirectedCycle dc = new DirectedCycle(g);
		if(!dc.hasCycle()){
			DepthFirstOrder dfo = new DepthFirstOrder(g);
			order = dfo.reversePost();	//depth first order的逆后序就是拓扑图的顺序。
		}
	}
	public Iterable<Integer> order(){
		return order;
	}
	public boolean isDAG(){
		return order != null;
	}
}

强连通分量

强连通分量

通过对强连通分量的研究可以将不同的对象进行分类，将具有相似性质的对象统一处理。

  • 接口
public interface SCC {
	/**
	 * @Description: Check if v and w are connected.
	 * @param v
	 * @param w
	 * @return
	 */
	public boolean strongConnected(int v, int w);
	/**
	 * @Description: The number of strong connected component.
	 * @return
	 */
	public int count();
	/**
	 * @Description: Which of the strong component belongs to.
	 * @param v
	 * @return
	 */
	int id(int v);
}
  • Kosaraju算法计算强连通分量
public class StrongCircleComponent implements SCC {
	private boolean[] marked;
	private int[] id;
	private int count;
	public StrongCircleComponent(Digraph g) {
		marked = new boolean[g.V()];
		id = new int[g.V()];
		DepthFirstOrder order = new DepthFirstOrder(g.reverse());//获得g的反转图
		for(int s : order.reversePost()){	//按照逆后序读取元素（拓扑图顺序）
			if(!marked[s]){
				dfs(g, s); count++;
			}
		}
	}
	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);
	}
	@Override
	public boolean strongConnected(int v, int w) {
		return id[v] == id[w];
	}
	@Override
	public int count() {
		return count;
	}
	@Override
	public int id(int v) {
		return id[v];
	}
}
  • Kosaraju证明
  1. 每次和s强连通的顶点都会在构造函数调用dfs(g, s)时被访问到。

反证法： 假设有一个和s强连通的顶点v不会被访问到，说明v曾经被访问过。 如果v曾经被访问过,s和v为强连通分量，s应被访问过。 s当前正在被访问，矛盾。

  1. 构造函数dfs(G,s)所到达的任意顶点v都必然是和s强连通的。

A和B强连通充要条件：A和B在图中连通，B和A在反向图中也连通。