[英]Path Compression , How does this code is path compression?
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package algorithm1;
/**
*
* @author Navin
*/
public class QuickUnionWeighted {
private int [] id;
private int [] size;
int numberOfChild;
int j=0;
public QuickUnionWeighted(int N){
id = new int[N];
size = new int[N];
for(int i=0;i<N;i++){
id[i] = i;
size[i]=1;
}
}
public int root(int i){
while (i != id[i]){
id[i] = id[id[i]];
i=id[i];
}
return i;
}
public boolean connected(int p,int q){
return(root(p) == root(q));
}
public void union(int p,int q){
int i = root(p);
int j = root(q);
// if(i == j) return;
if(size[i] < size[j]){
id[i] = j;
size[j] += size[i];
}else{
id[j] = i;
size[i] +=size[j];
}
for(int k=0;k<size.length;k++){
System.out.print(size[k]+" ");
}
}
public static void main(String [] args){
QuickUnionWeighted quw = new QuickUnionWeighted(10);
quw.union(3,0);
quw.union(4, 0);
quw.union(3, 5);
quw.union(3, 6);
quw.union(3,9);
}
}
Because as I examined the code, the id[i] = id[id[i]]
is pointing to the parent of the node, and the examined node is not moved to its grandparents and the tree is not flattened.因为当我检查代码时,
id[i] = id[id[i]]
指向节点的父节点,并且检查的节点没有移动到它的祖父节点,树也没有展平。
Kindly Guide.请指导。
id[i] = id[id[i]];
This line is the path compression.这一行是路径压缩。
id[i]
is the parent of node i
. id[i]
是节点i
的父节点。 Hence, this line re-links node i
to its grand-parent.因此,这条线将节点
i
重新链接到它的祖父节点。 Therefore, it skips the parent.因此,它跳过了父级。 Then, the same happens to the grand-parent and so on.
然后,祖父母等也会发生同样的情况。 This flattens the tree.
这会使树变平。
Here is a visualization of this step:这是这一步的可视化:
1 1
^ / \
| root(3) / \
2 --------> 2 3
^
|
3
I know this is long due.我知道这是迟到的。 But in case anyone still needs this.
但万一有人仍然需要这个。
Just as Nico said, this line is the path compression:正如 Nico 所说,这一行是路径压缩:
id[i] = id[id[i]];
But what it does is connect the node i
directly to its grand-parent.但它所做的是将节点
i
直接连接到它的祖父节点。 It then skips the parent node, and the whole process is repeated for the grandparent onwards.然后它跳过父节点,并且从祖父节点开始重复整个过程。 Thereby flattening the tree.
从而压平树。
However, to connect the node i
directly to the root node, here is another implementation of path compression:但是,为了将节点
i
直接连接到根节点,这里是路径压缩的另一种实现:
id[i] = root(id[i]);
This creates a recursive function that goes down to the root node and sets every node on that path to point directly to the root node.这将创建一个递归函数,该函数向下到根节点,并将该路径上的每个节点设置为直接指向根节点。
Check https://algorithms.tutorialhorizon.com/disjoint-set-union-find-algorithm-union-by-rank-and-path-compression/ and page 9 of this material https://cseweb.ucsd.edu/classes/sp15/cse100-ab/lectures/Lec17_pdf.pdf检查https://algorithms.tutorialhorizon.com/disjoint-set-union-find-algorithm-union-by-rank-and-path-compression/和本材料的第 9 页https://cseweb.ucsd.edu/classes /sp15/cse100-ab/lectures/Lec17_pdf.pdf
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