Friday, November 27, 2015

Graph Misc


http://segmentfault.com/a/1190000002680101

无向图的数据结构

Class Graph
private final int V;  边数
private int E; 边的数目
private Bag<Integer>[] adj; 邻接表,存储与该节点相邻的节点,一个链表数组

无向图的API

public class Graph
    Graph(int V)   创建一个含有V个节点但不含边的无向图
    Graph(In)    从输入流中读取一幅图
    int V()      返回图中有多少个节点
    int E()      边数
    addEdge(int v,int w)  添加一条边
    Iterable<Integer> adj(int v)   V节点相邻的所有顶点
    String toString       对象的字符串表示
public class Graph {
    private final int V;
    private int E;
    private Bag<Integer>[] adj;   //邻接表

    public Graph(int V){
        this.V = V;
        this.E = 0;
        adj = (Bag<Integer>[]) new Bag[V];
        for(int v = 0;v < V;v++){
            adj[v] = new Bag<Integer>();
        }
    }

    public Graph(In in){
        this(in.readInt());
        int E = in.readInt();
        for(int i = 0;i < E;i++){
            int v = in.readInt();
            int w = in.readInt();
            addEdge(v,w);
        }
    }

    public int V(){ return V;}
    public int E(){ return E;}

    public void addEdge(int v,int w){
        adj[v].add(w);
        adj[w].add(v);
        E++;
    }
    public Iterable<Integer>adj(int v){
        return adj[v];
    }

}
http://segmentfault.com/a/1190000002680168
在图中,我们很自然地会问这几个问题
  • 从一个顶点s能否到达顶点v?
  • 以s为顶点能到达的所有顶点?
解决能否到达问题的算法就是深度优先算法,使用深度优先算法获得的从s到v的路径的时间与路径的长度成正比。
//基于深度优先算法,搜索查找图中的路径
//解决单点路径问题,即,从一点开始是否存在到另一个点的路径
public class DepthFirstPaths {
    private boolean[] marked;//这个顶点调用过dfs了吗
    private int[] edgeTo;//从起点到该点路径上的最后一个顶点
    private final int s;//起点

    public DepthFirstPaths(Graph g,int s)
    {
        marked = new boolean[g.V()];
        edgeTo = new int[g.V()];
        this.s = s;
        dfs(g,s);
    }

    private void dfs(Graph G,int V){
        marked[V] = true;
        for(int w: G.adj(V)){
            if(!marked[w]){
                edgeTo[w] = V;//表示这条路径上,w之前是v
                dfs(G,w);
            }
        }
    }

    public boolean hasPathTo(int V){
        return marked[V];
    }

    public Iterable<Integer> pathTo(int V){
        if(!hasPathTo(V)) return null;
        Stack<Integer> path = new Stack<Integer>();
        for(int x = V;x != s; x = edgeTo[x])//从终点开始往上回溯
            path.push(x);
        path.push(s);
        return path;

    }
}
那还有一个重要的问题就是,“从s到v是否存在一条路径,如果有找出其中最短的那条。”最短路径问题
当然这路考虑的是每条边的都是权值为1的情况。
解决这个问题的算法就是广度优先搜索算法
下面给出其实现代码,其中的使用了一个队列用来保存需要遍历的顶点。其实很上一篇中的代码结构差不多,只不过遍历的顶点是依次从队列中取出的
public class BreadthFirstPaths {
    private boolean[] marked;
    private int[] edgeTo;
    private final int s;

    public BreadthFirstPaths(Graph G,int s)
    {
        marked = new boolean[G.V()];
        edgeTo = new int[G.V()];
        this.s = s;
        bfs(G,s);
    }
    private void bfs(Graph G,int s)
    {
        Queue<Integer> queue = new Queue<Integer>();
        marked[s] = true;
        queue.enqueue(s);
        while(!queue.isEmpty())
        {
            int v = queue.dequeue();//从队列中删除改顶点
            for(int w:G.adj(v))//对该顶点相邻的所有顶点进行遍历,标记为访问过,同时加入队列
            {
                edgeTo[w] = v;
                marked[w] = true;
                queue.enqueue(w);
            }
        }
    }
    public boolean hasPathTo(int v){
        return marked[v];
    }
    public Iterable<Integer> pathTo(int v)
    {
        Stack<Integer> path = new Stack<Integer>();
        while(edgeTo[v] != s)//同上一篇,通过该数组往上回溯
        {
            path.push(v);
            v = edgeTo[v];
        }
        path.push(s);
        return path;
    }
}
http://segmentfault.com/a/1190000002680241
连通分量是深度优先搜索算法的一个应用。
每进行了一次dfs,就会找到一条连通分量。
public class CC {
    /*
     * 计算一个图中的连通分量,从起始点开始进行dfs算法,同时用一个以顶点作为索引的数组id[]来保存该点所在的连通分量的起始点,也就是id
     * 这样,判断两个点是否处于同一个连通分量,只要判断id是否相同
     */
    private boolean[] marked;
    private int[] id;
    private int count;

    public CC(Graph G){
        marked = new boolean[G.V()];
        id = new int[G.V()];
        for(int s = 0;s < G.V();s++){
            if(!marked[s]){
                dfs(G,s);
                count++;
            }
        }
    }
    private void dfs(Graph G,int v){
        marked[v] = true;
        id[v] = count;          //该连通分量的起始节点
        for(int w:G.adj(v)){
            if(!marked[w])
                dfs(G,w);
        }
    }

    public boolean connected(int v,int w){
        return id[v] == id[w];
    }
    public int id(int v){
        return id[v];
    }
    public int count(){
        return count;
    }
}
同样还能解决双色问题,即“能够用两种不同颜色将顶点着色,使得相邻的顶点颜色不同吗?等价于这个图是一个二分图吗?
// not efficient implementation
/*
 * 使用dfs算法,来判断一个图中的顶点是否可以用两种颜色染色,使得相邻的顶点颜色不同
 * 也就是,判断该图是否是一个二分图:
 * 设G=(V,E)是一个无向图,如果顶点V可分割为两个互不相交的子集(A,B),并且图中的每条边(i,j)所关联的两个顶点i和j分别属于这两个不同的顶点集(i in A,j in B),则称图G为一个二分图。
 * 
 */
public class TwoColor {
    private boolean[] marked;
    private boolean[] color;
    private boolean isTwoColorable = true;

    public TwoColor(Graph G){
        marked = new boolean[G.V()];
        color = new boolean[G.V()];
        for(int s = 0;s < G.V();s++){
            if(!marked[s])
                dfs(G,s);
        }
    }

    private void dfs(Graph G,int v){
        marked[v] = true;
        for(int w: G.adj(v)){
            if(!marked[w]){
                color[w] =!color[v];
            }
            else if (color[w] == color[v])
                isTwoColorable = false;
        }
    }
    public boolean isTwoColorable(){
        return isTwoColorable;
    }
}

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