Tree_Graph 有向图是否存在路径 @CareerCup

原文：

Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

译文：

给定一个有向图，设计算法判断两结点间是否存在路径。

思路：

如果BFS就用队列

如果DFS就用递归回溯，参考回溯模板：http://blog.youkuaiyun.com/fightforyourdream/article/details/16997793

package Tree_Graph;

import java.util.LinkedList;

public class S4_2 {

	// ======================== State
	enum State{
		unvisited, visited, visiting;
	}
	
	// ======================== Node
	static class Node {
		private String vertex;
		private Node[] adjacent;
		public int adjacentCount;
		public State state;
		
		public Node(String vertex, int adjacentLength) {
			this.vertex = vertex;
			adjacentCount = 0;
			adjacent = new Node[adjacentLength];
		}
		
		public void addAdjacent(Node x) {
			if (adjacentCount < 30) {
				this.adjacent[adjacentCount++] = x;
			} else {
				System.out.println("No more adjacent can be added");
			}
		}
		
		public Node[] getAdjacent() {
			return adjacent;
		}
		
		public String getVertex() {
			return vertex;
		}
	}
	
	// ======================== Graph
	static class Graph {
		private Node[] vertices;
		public int count;
		
		public Graph() {
			vertices = new Node[6];
			count = 0;
		}
		
		public void addNode(Node x) {
			if (count < 30) {
				vertices[count] = x;
				count++;
			} else {
				System.out.println("Graph full");
			}
		}
		
		public Node[] getNodes() {
			return vertices;
		}
	}
	
	// ============================== BFS
	public static boolean bfs(Graph g, Node start, Node end) {
		LinkedList<Node> queue = new LinkedList<Node>();
		
		start.state = State.visiting;
		queue.add(start);
		while ( !queue.isEmpty() ) {
			Node cur = queue.remove();
			if (cur != null) {
				for (Node nbr : cur.getAdjacent()) {
					if (nbr.state == State.unvisited) {
						if (nbr == end) {
							return true;
						} else {
							nbr.state = State.visiting;
							queue.add(nbr);
						}
					}
				}
				cur.state = State.visited;
			}
		}
		return false;
	}
	
	// ================================= DFS
	public static boolean dfs(Graph g, Node start, Node end) {
		if ( start == end ) {
			return true;
		}
		
		for (Node nbr : start.getAdjacent()) {
			if (nbr.state == State.unvisited) {
				start.state = State.visiting;
				if(dfs(g, nbr, end)) {
					return true;
				}
				start.state = State.unvisited;
			}
		}
		return false;
	}
	
	
	
	public static Graph createNewGraph()
	{
		Graph g = new Graph();        
		Node[] temp = new Node[6];

		temp[0] = new Node("a", 3);
		temp[1] = new Node("b", 0);
		temp[2] = new Node("c", 0);
		temp[3] = new Node("d", 1);
		temp[4] = new Node("e", 1);
		temp[5] = new Node("f", 1);

		temp[0].addAdjacent(temp[1]);
		temp[0].addAdjacent(temp[2]);
		temp[0].addAdjacent(temp[3]);
		temp[3].addAdjacent(temp[4]);
		temp[4].addAdjacent(temp[5]);
		
		temp[5].addAdjacent(temp[0]);		// 添加一条边变成环
		
		for (int i = 0; i < 6; i++) {
			g.addNode(temp[i]);
		}
		resetState(g);
		return g;
	}
	
	public static void resetState(Graph g) {
		for (Node u : g.getNodes()) {
			u.state = State.unvisited;
		}
	}
	
	
	public static void main(String a[])
	{
		Graph g = createNewGraph();
		Node[] n = g.getNodes();
		Node start = n[4];
		Node end = n[3];
		resetState(g);
		System.out.println(bfs(g, start, end));
		resetState(g);
		System.out.println(dfs(g, start, end));
	}
	
	
}