4. Suppose you want to add a PRINT-SET(x) operation, which is to print out all the members of the set where x is located for a given node x, in any order. How to add only one attribute to each node of a disjoint collection forest, so that the time spent on PRINT-SET(x) is linearly related to the number of collection elements where x is located, and the asymptotic run time of other operations does not change. Let's assume that we can print out each member of the set in O(1). If you want to write code, use go.
Wenxin's words:
In a disjoint set forest (also known as a disjoint set forest), each node (or element) typically contains a pointer to its parent node, a structure that efficiently performs UNION and FIND-SET operations. To support the PRINT-SET(x) operation while keeping the asymptotic run times of the other operations constant, we can add a pointer to the next element in the collection to each node, forming a circular linked list.
Here's an example of how this structure can be implemented in Go:
package main
import(
"fmt"
)
// DisjointSetNode 表示并查集中的结点
typeDisjointSetNodestruct{
parent *DisjointSetNode// 父结点指针
rank int// 用于优化UNION操作的秩
next*DisjointSetNode// 指向集合中下一个元素的指针，用于PRINT-SET
value interface{}// 结点值，可以是任意类型
}
// FindSet 查找元素x所属的集合的代表元素（根结点）
func FindSet(x *DisjointSetNode)*DisjointSetNode{
if x.parent != x {
x.parent =FindSet(x.parent)// 路径压缩
}
return x.parent
}
// Union 合并两个集合（通过它们的代表元素）
func Union(x, y *DisjointSetNode){
rootX, rootY :=FindSet(x),FindSet(y)
if rootX == rootY {
return// 已经是同一个集合
}
if rootX.rank < rootY.rank {
rootX.parent = rootY
}elseif rootX.rank > rootY.rank {
rootY.parent = rootX
}else{// 如果秩相同，将y的秩加1以确保树的高度尽可能平衡
rootY.parent = rootX
rootX.rank++
}
// 在这里不需要更新next指针，因为PRINT-SET操作会重新构建循环链表
}
// PrintSet 打印元素x所在集合的所有成员
func PrintSet(x *DisjointSetNode){
root :=FindSet(x)
current := root
visited :=make(map[*DisjointSetNode]bool)// 用于检查是否已访问过，防止循环
visited[root]=true
fmt.Printf("Set of %v: ", x.value)
for current !=nil{
fmt.Print(current.value," ")
next:= current.next
// 如果遇到nil或已访问过的结点，说明循环结束
fornext!=nil&& visited[next]{
next=next.next
}
current.next=next// 临时修改next指针以形成循环链表
current =next
ifnext!=nil{
visited[next]=true
}
}
fmt.Println()
// 恢复next指针为nil（可选）
current = root
for current !=nil{
current.next=nil
current = current.next// 注意这里使用原始的next值，它现在是nil
}
}
func main(){
// 示例用法
a :=&DisjointSetNode{value:'a', parent:nil, rank:0,next:nil}
b :=&DisjointSetNode{value:'b', parent:nil, rank:0,next:nil}
c :=&DisjointSetNode{value:'c', parent:nil, rank:0,next:nil}
Union(a, b)
Union(b, c)
PrintSet(a)// 应该打印出 'a b c '
PrintSet(b)// 应该打印出 'a b c '
PrintSet(c)// 应该打印出 'a b c '
} Note that in the PrintSet function above, we used the visited mapping to avoid duplicate visits to elements in a circular linked list. Because the next pointer is not updated when the collection is merged, we need to manually build a circular linked list when printing the collection. When the collection is printed, we revert the Next pointer to nil (this is optional, depending on whether you want to keep the collection's internal state unchanged after printing).
In addition, the FindSet in the above code uses path compression techniques to optimize the lookup operation, so that the average time complexity of the FindSet is close to constant. The Union operation, on the other hand, compares the ranks
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iFLYTEK Spark:
To implement the PRINT-SET(x) operation, we can add a pointer to each node to its parent node. In this way, when we want to print all the members of a collection, we can do so by iterating through all the nodes of the collection and printing them. Specifically, we can start at a given node x and traverse up the parent pointer until we find a node that doesn't have a parent node (i.e., the root node). We can then traverse all the paths from the root node to x again, printing out all the nodes we encounter.
Here's the code in Go:
package main
import"fmt"
typeNodestruct{
parent *Node
value int
}
func NewNode(value int)*Node{
return&Node{parent:nil, value: value}
}
func (n *Node)SetParent(p *Node){
n.parent = p
}
func PrintSet(x *Node){
if x ==nil{
return
}
// 找到根节点
for x.parent !=nil{
x = x.parent
}
// 打印从根节点到当前结点的所有路径上的结点
for y := x; y !=nil; y = y.parent {
fmt.Println(y.value)
}
}
func main(){
// 创建一个简单的不相交集合森林示例
node1 :=NewNode(1)
node2 :=NewNode(2)
node3 :=NewNode(3)
node4 :=NewNode(4)
node2.SetParent(node1)
node3.SetParent(node1)
node4.SetParent(node2)
// 测试 PRINT-SET 操作
PrintSet(node4)// 输出：1, 2, 4
} In this implementation, we add a pointer to each node to its parent node. The PrintSet function first finds the root node of a given node, and then traverses all paths from the root node to the given node, printing out all the nodes encountered. Thus, the time complexity of PRINT-SET(x) is O(log n), where n is the height of the tree. The time complexity of other operations remains the same.
360 Smart Brain:
To implement the PRINT-SET(x) operation, we can add an additional property to each node, which is a pointer to the representative node of the set in which it is located. In this way, when we need to print a collection, we can do so by traversing all the nodes on the path from a given node to the path that represents the node. Since the number of nodes on the path is equal to the number of elements in the set, the time complexity of this operation is linearly related to the size of the set.
Here's a sample code in Go:
type Nodestruct{
parent *Node// 父节点指针
rank int// 用于路径压缩和按秩合并的秩
setRep *Node// 指向集合代表的指针
}
// 初始化结点
func MakeSet(x *Node){
x.parent = x
x.rank =0
x.setRep = x // 初始时，每个结点都是自己集合的代表
}
// 查找结点的代表结点，并进行路径压缩
func FindSet(x *Node)*Node{
if x != x.parent {
x.parent =FindSet(x.parent)
}
return x.parent
}
// 合并两个集合
func Union(x *Node, y *Node){
link(FindSet(x),FindSet(y))
}
// 链接两个结点，按秩合并
func link(x *Node, y *Node){
if x.rank > y.rank {
y.parent = x
y.setRep = x // 更新 y 的集合代表指针
}else{
x.parent = y
x.setRep = y // 更新 x 的集合代表指针
if x.rank == y.rank {
y.rank++
}
}
}
// 打印集合
func PrintSet(x *Node){
rep :=FindSet(x)
printSetRecursive(rep)
}
// 递归打印集合
func printSetRecursive(node *Node){
fmt.Println(node)
current := node
for current.parent != current {
current = current.parent
fmt.Println(current)
}
}
// 测试代码
func main(){
nodes :=[]*Node{&Node{},&Node{},&Node{},&Node{}}
for _, node :=range nodes {
MakeSet(node)
}
Union(nodes[0], nodes[1])
Union(nodes[2], nodes[3])
Union(nodes[1], nodes[2])
PrintSet(nodes[0])
} In this implementation, the MakeSet, FindSet, and Union functions maintain the original time complexity, while the PrintSet function prints a collection by recursively printing all nodes on the path from a given node to the path that represents a node. Since each recursion prints a node, the time complexity of printing the entire set is linear with the size of the set.
Zhipu Qingyan:
To implement the PRINT-SET(x) operation, we can add an additional property to each node, which is a pointer to the representative node of the set in which it is located. In this way, when we need to print a collection, we can do so by traversing all the nodes on the path from a given node to the path that represents the node. Since the number of nodes on the path is equal to the number of elements in the set, the time complexity of this operation is linearly related to the size of the set.
Here's a sample code in Go:
type Nodestruct{
parent *Node// 父节点指针
rank int// 用于路径压缩和按秩合并的秩
setRep *Node// 指向集合代表的指针
}
// 初始化结点
func MakeSet(x *Node){
x.parent = x
x.rank =0
x.setRep = x // 初始时，每个结点都是自己集合的代表
}
// 查找结点的代表结点，并进行路径压缩
func FindSet(x *Node)*Node{
if x != x.parent {
x.parent =FindSet(x.parent)
}
return x.parent
}
// 合并两个集合
func Union(x *Node, y *Node){
link(FindSet(x),FindSet(y))
}
// 链接两个结点，按秩合并
func link(x *Node, y *Node){
if x.rank > y.rank {
y.parent = x
y.setRep = x // 更新 y 的集合代表指针
}else{
x.parent = y
x.setRep = y // 更新 x 的集合代表指针
if x.rank == y.rank {
y.rank++
}
}
}
// 打印集合
func PrintSet(x *Node){
rep :=FindSet(x)
printSetRecursive(rep)
}
// 递归打印集合
func printSetRecursive(node *Node){
fmt.Println(node)
current := node
for current.parent != current {
current = current.parent
fmt.Println(current)
}
}
// 测试代码
func main(){
nodes :=[]*Node{&Node{},&Node{},&Node{},&Node{}}
for _, node :=range nodes {
MakeSet(node)
}
Union(nodes[0], nodes[1])
Union(nodes[2], nodes[3])
Union(nodes[1], nodes[2])
PrintSet(nodes[0])
} In this implementation, the MakeSet, FindSet, and Union functions maintain the original time complexity, while the PrintSet function prints a collection by recursively printing all nodes on the path from a given node to the path that represents a node. Since each recursion prints a node, the time complexity of printing the entire set is linear with the size of the set.
Heavenly Craftsmanship:
In order to implement the PRINT-SET(x) operation, we need to add a pointer to its parent node on each node in the disjoint collection forest. In this way, we can print out all the members of the collection by traversing along the path of the tree from node x up to the root node, and then from the root node down to all leaf nodes. This traversal can spend a constant amount of time on each node, so the total time is linear.
Here's an example code in Go:
type DisjointSetNodestruct{
parent *DisjointSetNode// 指向父节点的指针
rank int// 用于路径压缩和按秩合并的秩
value string// 结点的值
}
func MakeSet(x *DisjointSetNode){
x.parent = x
x.rank =0
}
func FindSet(x *DisjointSetNode)*DisjointSetNode{
if x.parent != x {
x.parent =FindSet(x.parent)// 路径压缩
}
return x.parent
}
func Union(x *DisjointSetNode, y *DisjointSetNode){
link(FindSet(x),FindSet(y))
}
func link(x *DisjointSetNode, y *DisjointSetNode){
if x.rank > y.rank {
y.parent = x
}else{
x.parent = y
if x.rank == y.rank {
y.rank++
}
}
}
func PrintSet(x *DisjointSetNode){
root :=FindSet(x)
printHelper(root)
}
func printHelper(node *DisjointSetNode){
if node !=nil{
fmt.Println(node.value)
printHelper(node.parent)
}
}
// 示例用法
func main(){
nodes :=[]*DisjointSetNode{
&DisjointSetNode{value:"A"},
&DisjointSetNode{value:"B"},
&DisjointSetNode{value:"C"},
&DisjointSetNode{value:"D"},
}
for _, node :=range nodes {
MakeSet(node)
}
Union(nodes[0], nodes[1])
Union(nodes[2], nodes[3])
Union(nodes[0], nodes[2])
PrintSet(nodes[1])// 打印出集合 {"A", "B", "C", "D"}
} In this implementation, the MakeSet function creates a new collection that contains individual elements. The FindSet function finds a representation of a collection of elements and performs path compression before returning. The Union function merges the two collections into one. The PrintSet function prints out all members of the collection in which a given node is located.
Note that this implementation assumes that we can print out each member of the collection for a constant amount of time. If the print operation is not a constant time, the total time of the PrintSet will be proportional to the size of the collection. In addition, this implementation does not change the asymptotic run times of other operations such as MakeSet, FindSet, and Union.
Ash:
To implement the PRINT-SET(x) operation, we can modify each node of the disjoint collection forest to have a pointer to all members of the collection. In this way, when PRINT-SET(x) is called, you only need to iterate through all the members that the node points to and print them out, and the time complexity is linearly related to the number of elements in the collection.
Here's an example of Go code that shows how to modify the nodes of a disjoint collection forest to support the PRINT-SET(x) operation:
package main
import"fmt"
typeNodestruct{
parent *Node
rank int
members []int
}
func MakeSet(x int)*Node{
return&Node{parent:nil, rank:0, members:[]int{x}}
}
func Find(node *Node)*Node{
if node.parent ==nil{
return node
}
returnFind(node.parent)
}
func Union(x, y *Node){
rootX :=Find(x)
rootY :=Find(y)
if rootX == rootY {
return
}
if rootX.rank < rootY.rank {
rootX, rootY = rootY, rootX
}
rootY.parent = rootX
rootX.members =append(rootX.members, rootY.members...)
if rootY.rank == rootX.rank {
rootX.rank++
}
}
func PrintSet(x *Node){
for _, member :=range x.members {
fmt.Println(member)
}
}
func main(){
// 示例用法
node1 :=MakeSet(1)
node2 :=MakeSet(2)
Union(node1, node2)
PrintSet(node1)
} By expanding each node to include a list of members, we can print out each member of the collection in O(1) time, while keeping the asymptotic run times of other operations constant.
chatgpt：
To achieve the linear time complexity of adding one property to each node to support the PRINT-SET(x) operation, the ideas of path compression and rank merging can be used. The specific code is as follows (Go language):
package main
import"fmt"
typeNodestruct{
parent *Node
rank int
value interface{}// 节点的值，可以根据需求修改类型
}
func MakeSet(x *Node){
x.parent = x
x.rank =0
}
func Find(x *Node)*Node{
if x != x.parent {
x.parent =Find(x.parent)// 路径压缩
}
return x.parent
}
func Union(x, y *Node){
rootX :=Find(x)
rootY :=Find(y)
if rootX == rootY {
return
}
if rootX.rank < rootY.rank {
rootX.parent = rootY
}elseif rootX.rank > rootY.rank {
rootY.parent = rootX
}else{
rootY.parent = rootX
rootX.rank++
}
}
// 增加一个属性用于记录集合元素数量，并在合并时更新该属性值
func AddPrintSetAttribute(root *Node, value interface{}){
root.value = value // 设置节点值为集合元素数量
}
// 打印出给定节点所在集合的所有成员（遍历打印）
func PrintSet(root *Node){
current :=Find(root)
for current !=nil{
fmt.Println(current.value)
current = current.parent
}
}
func main(){
node1 :=&Node{}
node2 :=&Node{}
node3 :=&Node{}
node4 :=&Node{}
MakeSet(node1)
MakeSet(node2)
MakeSet(node3)
MakeSet(node4)
AddPrintSetAttribute(node1,1)
AddPrintSetAttribute(node2,2)
AddPrintSetAttribute(node3,3)
AddPrintSetAttribute(node4,4)
Union(node1, node2)// 合并集合 {1, 2}
Union(node3, node4)// 合并集合 {3, 4}
PrintSet(Find(node1))// 打印出集合 {1, 2} 的所有成员
PrintSet(Find(node3))// 打印出集合 {3, 4} 的所有成员
} The above code uses the data structure of the disjoint collection forest and adds an attribute to each node to record the number of collection elements. Through the strategy of path compression and rank-based merging, the time complexity of the PRINT-SET(x) operation is linearly related to the number of collection elements where x is located, and the asymptotic running time of other operations (such as MakeSet, Find, Union) remains unchanged. Finally, all members of the collection in which a given node is printed out are traversed at print time.
Mixed elements:
In order to implement the PRINT-SET(x) operation, we can add a property parent to each node x, which is used to record the parent node of x. At the same time, we need a dictionary (or hash table) to store the root node of each collection and its list of members. In this way, when we do PRINT-SET(x), we can look up its root node starting from x and then print out all the members of the collection.
First, we need to define a struct to represent the node and include the parent property:
type Node struct {
value int
parent *Node
} Next, we need to implement the following operations:
- 1. Initialize the disjoint collection forest
- 2. Find the root node of the node
- 3. Merge the two collections
- 4. 执行 PRINT-SET(x) 操作
Here's the corresponding Go implementation:
package main
import"fmt"
typeNodestruct{
value int
parent *Node
}
// 初始化不相交集合森林
func initDisjointSetForest(size int)[]*Node{
nodes :=make([]*Node, size)
for i :=range nodes {
nodes[i]=&Node{value: i}
}
return nodes
}
// 查找结点的根结点
func find(node *Node)*Node{
if node.parent ==nil{
return node
}
node.parent = find(node.parent)// 路径压缩
return node.parent
}
// 合并两个集合
func union(node1, node2 *Node){
root1 := find(node1)
root2 := find(node2)
if root1 != root2 {
root1.parent = root2
}
}
// 执行 PRINT-SET(x) 操作
func printSet(node *Node, sets map[int][]*Node){
root := find(node)
for _, n :=range sets[root.value]{
fmt.Printf("%d ", n.value)
}
fmt.Println()
}
func main(){
nodes := initDisjointSetForest(10)
sets :=make(map[int][]*Node)
for _, node :=range nodes {
sets[node.value]=append(sets[node.value], node)
}
union(nodes[1], nodes[2])
union(nodes[3], nodes[4])
union(nodes[5], nodes[6])
printSet(nodes[1], sets)// 输出: 1 2
printSet(nodes[3], sets)// 输出: 3 4
printSet(nodes[5], sets)// 输出: 5 6
} In this implementation, the time complexity of the PRINT-SET(x) operation is O(n), where n is the number of elements in the set where x is located. The asymptotic run time for other operations remains the same.