我正在尝试实现并查算法,但我查找的所有实现都使用整数。我需要实现该算法,以便可以通过以下方式调用 union() 和connected() 方法: union(Vertex v, Vertex, w) -connected(Vertex v, Vertex w)
我已经尝试调整我的算法以使其能够处理顶点,但我不知道如何替换父属性和排名属性以使其工作。请帮忙:(
公开课用友{
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));
}
}
}
最佳答案
在标准算法中,每个顶点都会被赋予一个int
id,它表示它在数组中的位置。所以这意味着 parent[0]
包含顶点 0 的父级的 id 等等。
实际上,您可以将数组视为从 int
到其他内容的非常有效的映射。如果您将 int
替换为更复杂的类型,那么您需要开始使用 Map
而不是数组。
因此,如果您想使用名为 Vertex
的类来表示顶点,那么您需要以不同的方式声明父级和等级:
Map<Vertex,Vertex> parent = new HashMap<>();
Map<Vertex,Rank> rank = new HashMap<>();
如果您想坚持当前方案,您可以将 Rank
替换为 Byte
- 尽管使用类可能是更好的封装。
然后您将得到如下所示的代码:
while (!vertex.equals(parent.get(vertex))) {
parent.put(vertex, parent.get(parent.get(vertex)));
vertex = parent.get(vertex);
}
return vertex;
需要注意的一件事是,如果您要使用 Vertex
作为 map 的键(正如我所建议的),那么您必须实现equals
和 hashCode
方法。
关于java - 如何实现并查算法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56230697/