Question

我正在尝试实现Union-Find算法，但是我查询的所有实现都使用整数。我需要实现算法，以便可以通过这种方式调用union（）和connected（）方法：union（Vertex v，Vertex，w）-connected（Vertex v，Vertex w）

我已经尝试过调整算法使其适用于顶点，但是我不知道如何替换父代和对属性进行排序以使其起作用。请帮助：（

公共类UF {

private int[] parent;  // parent[i] = parent of i
private byte[] rank;   // rank[i] = rank of subtree rooted at i (never more than 31)
private int count;     // number of components

/**
 * Initializes an empty union–find data structure with {@code n} sites
 * {@code 0} through {@code n-1}. Each site is initially in its own 
 * component.
 *
 * @param  n the number of sites
 * @throws IllegalArgumentException if {@code n < 0}
 */
public UF(int n) {
    if (n < 0) throw new IllegalArgumentException();
    count = n;
    parent = new int[n];
    rank = new byte[n];
    for (int i = 0; i < n; i++) {
        parent[i] = i;
        rank[i] = 0;
    }
}

/**
 * Returns the component identifier for the component containing site {@code p}.
 *
 * @param  p the integer representing one site
 * @return the component identifier for the component containing site {@code p}
 * @throws IllegalArgumentException unless {@code 0 <= p < n}
 */
public int find(int p) {
    validate(p);
    while (p != parent[p]) {
        parent[p] = parent[parent[p]];    // path compression by halving
        p = parent[p];
    }
    return p;
}

/**
 * Returns the number of components.
 *
 * @return the number of components (between {@code 1} and {@code n})
 */
public int count() {
    return count;
}

/**
 * Returns true if the the two sites are in the same component.
 *
 * @param  p the integer representing one site
 * @param  q the integer representing the other site
 * @return {@code true} if the two sites {@code p} and {@code q} are in the same component;
 *         {@code false} otherwise
 * @throws IllegalArgumentException unless
 *         both {@code 0 <= p < n} and {@code 0 <= q < n}
 */
public boolean connected(int p, int q) {
    return find(p) == find(q);
}

/**
 * Merges the component containing site {@code p} with the 
 * the component containing site {@code q}.
 *
 * @param  p the integer representing one site
 * @param  q the integer representing the other site
 * @throws IllegalArgumentException unless
 *         both {@code 0 <= p < n} and {@code 0 <= q < n}
 */
public void union(int p, int q) {
    int rootP = find(p);
    int rootQ = find(q);
    if (rootP == rootQ) return;

    // make root of smaller rank point to root of larger rank
    if      (rank[rootP] < rank[rootQ]) parent[rootP] = rootQ;
    else if (rank[rootP] > rank[rootQ]) parent[rootQ] = rootP;
    else {
        parent[rootQ] = rootP;
        rank[rootP]++;
    }
    count--;
}

// validate that p is a valid index
private void validate(int p) {
    int n = parent.length;
    if (p < 0 || p >= n) {
        throw new IllegalArgumentException("index " + p + " is not between 0 and " + (n-1));  
    }
}

}

Answer 1

在标准算法中，每个顶点都有一个int id表示它在数组中的位置。因此，这意味着parent[0]包含顶点0的父代的ID，以此类推。

真的，您可以将数组视为从int到其他对象的非常有效的映射。如果将int替换为更复杂的类型，则需要开始使用Map而不是数组。

因此，如果您想使用名为Vertex的类来表示顶点，则需要声明父对象并以不同的顺序排列：

Map<Vertex,Vertex> parent = new HashMap<>();
Map<Vertex,Rank> rank = new HashMap<>();

如果要坚持使用当前方案，可以将Rank替换为Byte-尽管使用类最好封装。

然后您将得到类似于以下内容的代码：

while (!vertex.equals(parent.get(vertex))) {
    parent.put(vertex, parent.get(parent.get(vertex)));
    vertex = parent.get(vertex);
}
return vertex;

要注意的一件事是，如果您打算将Vertex用作地图的键（如我所建议的那样），那么您必须实现{{1} }和equals方法。

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1 个答案: