Question

我目前难以从哪里开始做家庭作业。基本上，我们已经获得了一个函数和一些广度优先搜索代码。目标是计算上面列出的东西，他强烈建议使用BFS代码来实现目标。任何帮助都表示赞赏：

package algs41;

import stdlib.*;

// This is problem 4.1.16 from the textbook
//
// You need only change the constructor.
// You can also change the main method.
// But you should not change eccentricity(), diameter(), radius(), center() or isCenter()
// You can (and should!) add more private methods (such as bfs or dfs)
public class MyGraphProperties {
    int[] eccentricity;
    int diameter;
    int radius;
    int center; 

    // Constructor can be linear in G.V() * G.E()
    public MyGraphProperties(Graph G) {
        this.eccentricity = new int[G.V()];
        int diameter = Integer.MIN_VALUE;
        int radius = Integer.MAX_VALUE;
        int center = -1;
        // If G.V()==0, then these are the correct values.
        // If G is not connected, you should throw a new IllegalArgumentException()
        // I suggest that you first get it to work for a connected graph
        // This will require that you traverse the graph starting from some node (say 0)
        // You can then adjust your code so that it throws an exception in the case that all nodes are not visited

        // TODO
        // compute the eccentricity, diameter, radius and center
        // The center of the graph is not unique, set center to SOME center --- it does not matter which one
        this.diameter = diameter;
        this.radius   = radius;
        this.center   = center;
    }

    // Do not change the following constant time methods
    public int eccentricity(int v) { return eccentricity[v]; }
    public int diameter()          { return diameter; }
    public int radius()            { return radius; }
    public int center()            { return center; }
    public boolean isCenter(int v) { return eccentricity[v] == radius; }

以下是相关的BFS代码：

    private void bfs(Graph G, int s) {
        Queue<Integer> q = new Queue<>();
        for (int v = 0; v < G.V(); v++) distTo[v] = INFINITY;
        distTo[s] = 0;
        marked[s] = true;
        q.enqueue(s);

        while (!q.isEmpty()) {
            int v = q.dequeue();
            for (int w : G.adj(v)) {
                if (!marked[w]) {
                    edgeTo[w] = v;
                    distTo[w] = distTo[v] + 1;
                    marked[w] = true;
                    q.enqueue(w);
                }
            }
        }
    }

目前不存在某些事物（例如distTo）。这是因为不允许使用我们给出的BreadthFirstSearch类，但我们可以复制并通过相关代码供我们自己使用。 请尽可能少地编辑bfs void。根据教授的说法，我所提供的bfs就是我们所需要的如上所述，我不确定从哪里开始。任何帮助表示赞赏。

编辑：添加了图表代码

public class Graph {
private final int V;
private int E;
private final Bag<Integer>[] adj;

/**
 * Create an empty graph with V vertices.
 */
@SuppressWarnings("unchecked")
public Graph(int V) {
    if (V < 0) throw new Error("Number of vertices must be nonnegative");
    this.V = V;
    this.E = 0;
    this.adj = new Bag[V];
    for (int v = 0; v < V; v++) {
        adj[v] = new Bag<>();
    }
}

/**
 * Create a random graph with V vertices and E edges with no parallel edges or self loops.
 * Expected running time is proportional to V + E.
 */
public Graph(int V, int E) { this (V, E, false); }
/**
 * Create a random graph with V vertices and E edges.
 * Expected running time is proportional to V + E.
 */
public Graph(int V, int E, boolean allowParallelEdgesAndSelfLoops) {
    this(V);
    if (E < 0) throw new Error("Number of edges must be nonnegative");
    if (allowParallelEdgesAndSelfLoops) {
        for (int i = 0; i < E; i++) {
            int v = (int) (Math.random() * V);
            int w = (int) (Math.random() * V);
            addEdge(v, w);
        }
    } else {
        if (E > V*(V-1)/2) throw new Error("Number of edges must be less than V*(V-1)/2");
        newEdge: while (E>0) {
            int v = (int) (Math.random() * V);
            int w = (int) (Math.random() * V);
            if (v == w) continue;
            for (int w2 : adj[v])
                if (w == w2)
                    continue newEdge;
            addEdge(v, w);
            E--;
        }
    }
}

/**
 * Create a digraph from input stream.
 */
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);
    }
}

/**
 * Copy constructor.
 */
public Graph(Graph G) {
    this(G.V());
    this.E = G.E();
    for (int v = 0; v < G.V(); v++) {
        // reverse so that adjacency list is in same order as original
        Stack<Integer> reverse = new Stack<>();
        for (int w : G.adj[v]) {
            reverse.push(w);
        }
        for (int w : reverse) {
            adj[v].add(w);
        }
    }
}

/**
 * Return the number of vertices in the graph.
 */
public int V() { return V; }

/**
 * Return the number of edges in the graph.
 */
public int E() { return E; }


/**
 * Add the undirected edge v-w to graph.
 * @throws java.lang.IndexOutOfBoundsException unless both 0 <= v < V and 0 <= w < V
 */
public void addEdge(int v, int w) {
    if (v < 0 || v >= V) throw new IndexOutOfBoundsException();
    if (w < 0 || w >= V) throw new IndexOutOfBoundsException();
    E++;
    adj[v].add(w);
    adj[w].add(v);
}


/**
 * Return the list of neighbors of vertex v as in Iterable.
 * @throws java.lang.IndexOutOfBoundsException unless 0 <= v < V
 */
public Iterable<Integer> adj(int v) {
    if (v < 0 || v >= V) throw new IndexOutOfBoundsException();
    return adj[v];
}

enter image description here

计算偏心率，直径，半径和中心（使用Java代码）

0 个答案: