查找给定查询中每个元素所属的数组以及元素计数

给定一个长度为N的数组对arr[][]和一个长度为M的数组queries[]和一个整数R ，其中每个查询包含一个从1到R的整数，每个查询[i]的任务是找到它所属的集合，并找到该集合的元素总数。

Note: Initially every integer from 1 to R belongs to the distinct set.

例子：

Input: R = 5, arr[] = {{1, 2}, {2, 3}, {4, 5}}, queries[] = {2, 4, 1, 3}
Output: 3 2 3 3
Explanation:
After making the sets from the arr[] pairs, {1, 2, 3}, {4, 5}
For the first query: 2 belongs to the set {1, 2, 3} and the total number of elements is 3.
For the second query: 4 belongs to the set {4, 5} and the total number of elements is 2.
For the third query: 1 belongs to the set {1, 2, 3} and the total number of elements is 3.
For the fourth query: 3 belongs to the set {1, 2, 3} and the total number of elements is 3.

Input: R = 6, arr[] = {{1, 3}, {2, 4}}, queries[] = {2, 5, 6, 1}
Output: 2 1 1 2

方法：给定的问题可以使用不相交集并集来解决。最初，所有元素都在不同的集合中，处理arr[]并对给定的对进行联合操作，并在联合更新中，父元素的总 []值。对于每个查询执行查找操作，并为返回的父元素查找当前集合的大小作为total[parent]的值。请按照以下步骤解决问题：

• 初始化向量parent(R + 1) 、 rank(R + 1, 0) 、 total(R + 1, 1) 。
• 使用变量i遍历范围[1, R+1)并将parent[I]的值设置为I 。
• 使用变量i遍历范围[1, N-1]并执行联合运算作为Union(parent, rank, total, arr[I].first, arr[I].second) 。
• 使用变量i迭代范围[1, M – 1]并执行以下步骤：
  • 调用函数来查找当前元素的父元素queries[i]为Find(parent, queries[I]) 。
  • 将total[i]的值打印为当前集合的大小。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to perform the find operation
// of disjoint set union
int Find(vector& parent, int a)
{
    return parent[a]
           = (parent[a] == a)
                 ? a
                 : (Find(parent, parent[a]));
}
 
// Function to find the Union operation
// of disjoint set union
void Union(vector& parent,
           vector& rank,
           vector& total,
           int a, int b)
{
    // Find the parent of a and b
    a = Find(parent, a);
    b = Find(parent, b);
 
    if (a == b)
        return;
 
    // If the rank are the same
    if (rank[a] == rank[b]) {
        rank[a]++;
    }
 
    if (rank[a] < rank[b]) {
        int temp = a;
        a = b;
        b = temp;
    }
 
    // Update the parent for node b
    parent[b] = a;
 
    // Update the total number of
    // elements of a
    total[a] += total[b];
}
 
// Function to find the total element
// of the set which belongs to the
// element queries[i]
void findTotNumOfSet(vector >& arr,
                     vector& queries,
                     int R, int N, int M)
{
 
    // Stores the parent elements
    // of the sets
    vector parent(R + 1);
 
    // Stores the rank of the sets
    vector rank(R + 1, 0);
 
    // Stores the total number of
    // elements of the sets
    vector total(R + 1, 1);
 
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] to i
        parent[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add the arr[i].first and
        // arr[i].second elements to
        // the same set
        Union(parent, rank, total,
              arr[i].first,
              arr[i].second);
    }
 
    for (int i = 0; i < M; i++) {
 
        // Find the parent element of
        // the element queries[i]
        int P = Find(parent, queries[i]);
 
        // Print the total elements of
        // the set which belongs to P
        cout << total[P] << " ";
    }
}
 
// Driver Code
int main()
{
    int R = 5;
    vector > arr{ { 1, 2 },
                                 { 2, 3 },
                                 { 4, 5 } };
    vector queries{ 2, 4, 1, 3 };
    int N = arr.size();
    int M = queries.size();
 
    findTotNumOfSet(arr, queries, R, N, M);
 
    return 0;
}

Java

// Java program for the above approach
class GFG
{
 
// Function to perform the find operation
// of disjoint set union
static int Find(int [] parent, int a)
{
    return parent[a]
           = (parent[a] == a)
                 ? a
                 : (Find(parent, parent[a]));
}
 
// Function to find the Union operation
// of disjoint set union
static void Union(int [] parent,
           int [] rank,
           int [] total,
           int a, int b)
{
    // Find the parent of a and b
    a = Find(parent, a);
    b = Find(parent, b);
 
    if (a == b)
        return;
 
    // If the rank are the same
    if (rank[a] == rank[b]) {
        rank[a]++;
    }
 
    if (rank[a] < rank[b]) {
        int temp = a;
        a = b;
        b = temp;
    }
 
    // Update the parent for node b
    parent[b] = a;
 
    // Update the total number of
    // elements of a
    total[a] += total[b];
}
 
// Function to find the total element
// of the set which belongs to the
// element queries[i]
static void findTotNumOfSet(int[][] arr,
                     int [] queries,
                     int R, int N, int M)
{
 
    // Stores the parent elements
    // of the sets
    int [] parent = new int[R + 1];
 
    // Stores the rank of the sets
    int [] rank = new int[R + 1];
 
    // Stores the total number of
    // elements of the sets
    int [] total = new int[R + 1];
    for (int i = 0; i < total.length; i++) {
        total[i] = 1;
    }
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] to i
        parent[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add the arr[i][0] and
        // arr[i][1] elements to
        // the same set
        Union(parent, rank, total,
              arr[i][0],
              arr[i][1]);
    }
 
    for (int i = 0; i < M; i++) {
 
        // Find the parent element of
        // the element queries[i]
        int P = Find(parent, queries[i]);
 
        // Print the total elements of
        // the set which belongs to P
        System.out.print(total[P]+ " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int R = 5;
    int[][]  arr = { { 1, 2 },
                                 { 2, 3 },
                                 { 4, 5 } };
    int [] queries = { 2, 4, 1, 3 };
    int N = arr.length;
    int M = queries.length;
 
    findTotNumOfSet(arr, queries, R, N, M);
 
}
}
 
// This code is contributed by 29AjayKumar.

Python3

# Python3 program for the above approach
 
# Function to perform the find operation
# of disjoint set union
def Find(parent, a):
    if (parent[a] == a):
        return a
    else:
        return Find(parent, parent[a])
 
# Function to find the Union operation
# of disjoint set union
def Union(parent, rank, total, a, b):
   
    # Find the parent of a and b
    a = Find(parent, a)
    b = Find(parent, b)
    if(a == b):
        return
       
    # If the rank are the same
    if(rank[a] == rank[b]):
        rank[a] += 1
    if(rank[a] < rank[b]):
        temp = a
        a = b
        b = temp
         
    # Update the parent for node b
    parent[b] = a
     
    # Update the total number of
    # elements of a
    total[a] += total[b]
 
# Function to find the total element
# of the set which belongs to the
# element queries[i]
def findTotNumOfSet(arr, queries, R, N, M):
   
    # Stores the parent elements
    # of the sets
    parent = [None]*(R+1)
     
    # Stores the rank of the sets
    rank = [0]*(R+1)
     
    # Stores the total number of
    # elements of the sets
    total = [1]*(R + 1)
    for i in range(1, R + 1):
       
        # Add the arr[i].first and
        # arr[i].second elements to
        # the same set
        parent[i] = i
    for i in range(N):
        Union(parent, rank, total, arr[i][0], arr[i][1])
    for i in range(M):
       
        # Find the parent element of
        # the element queries[i]
        P = Find(parent, queries[i])
         
        # Print the total elements of
        # the set which belongs to P
        print(total[P], end=" ")
 
# Driver code
R = 5
arr = [[1, 2], [2, 3], [4, 5]]
queries = [2, 4, 1, 3]
N = len(arr)
M = len(queries)
findTotNumOfSet(arr, queries, R, N, M)
 
# This code is contributed by parthmanchanda81

C#

// C# program for the above approach
using System;
 
public class GFG
{
 
// Function to perform the find operation
// of disjoint set union
static int Find(int [] parent, int a)
{
    return parent[a]
           = (parent[a] == a)
                 ? a
                 : (Find(parent, parent[a]));
}
 
// Function to find the Union operation
// of disjoint set union
static void Union(int [] parent,
           int [] rank,
           int [] total,
           int a, int b)
{
    // Find the parent of a and b
    a = Find(parent, a);
    b = Find(parent, b);
 
    if (a == b)
        return;
 
    // If the rank are the same
    if (rank[a] == rank[b]) {
        rank[a]++;
    }
 
    if (rank[a] < rank[b]) {
        int temp = a;
        a = b;
        b = temp;
    }
 
    // Update the parent for node b
    parent[b] = a;
 
    // Update the total number of
    // elements of a
    total[a] += total[b];
}
 
// Function to find the total element
// of the set which belongs to the
// element queries[i]
static void findTotNumOfSet(int[,] arr,
                     int [] queries,
                     int R, int N, int M)
{
 
    // Stores the parent elements
    // of the sets
    int [] parent = new int[R + 1];
 
    // Stores the rank of the sets
    int [] rank = new int[R + 1];
 
    // Stores the total number of
    // elements of the sets
    int [] total = new int[R + 1];
    for (int i = 0; i < total.Length; i++) {
        total[i] = 1;
    }
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] to i
        parent[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add the arr[i,0] and
        // arr[i,1] elements to
        // the same set
        Union(parent, rank, total,
              arr[i,0],
              arr[i,1]);
    }
 
    for (int i = 0; i < M; i++) {
 
        // Find the parent element of
        // the element queries[i]
        int P = Find(parent, queries[i]);
 
        // Print the total elements of
        // the set which belongs to P
        Console.Write(total[P]+ " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int R = 5;
    int[,]  arr = { { 1, 2 },
                                 { 2, 3 },
                                 { 4, 5 } };
    int [] queries = { 2, 4, 1, 3 };
    int N = arr.GetLength(0);
    int M = queries.GetLength(0);
 
    findTotNumOfSet(arr, queries, R, N, M);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

// Javascript program for the above approach
 
// Function to perform the find operation
// of disjoint set union
function Find(parent, a)
{
  return (parent[a] = parent[a] == a ? a : Find(parent, parent[a]));
}
 
// Function to find the Union operation
// of disjoint set union
function Union(parent, rank, total, a, b)
{
 
  // Find the parent of a and b
  a = Find(parent, a);
  b = Find(parent, b);
 
  if (a == b) return;
 
  // If the rank are the same
  if (rank[a] == rank[b]) {
    rank[a]++;
  }
 
  if (rank[a] < rank[b]) {
    let temp = a;
    a = b;
    b = temp;
  }
 
  // Update the parent for node b
  parent[b] = a;
 
  // Update the total number of
  // elements of a
  total[a] += total[b];
}
 
// Function to find the total element
// of the set which belongs to the
// element queries[i]
function findTotNumOfSet(arr, queries, R, N, M)
{
 
  // Stores the parent elements
  // of the sets
  let parent = new Array(R + 1);
 
  // Stores the rank of the sets
  let rank = new Array(R + 1).fill(0);
 
  // Stores the total number of
  // elements of the sets
  let total = new Array(R + 1).fill(1);
 
  for (let i = 1; i < R + 1; i++)
  {
   
    // Update parent[i] to i
    parent[i] = i;
  }
 
  for (let i = 0; i < N; i++)
  {
   
    // Add the arr[i].first and
    // arr[i].second elements to
    // the same set
    Union(parent, rank, total, arr[i][0], arr[i][1]);
  }
 
  for (let i = 0; i < M; i++)
  {
   
    // Find the parent element of
    // the element queries[i]
    let P = Find(parent, queries[i]);
 
    // Print the total elements of
    // the set which belongs to P
    document.write(total[P] + " ");
  }
}
 
// Driver Code
let R = 5;
let arr = [
  [1, 2],
  [2, 3],
  [4, 5],
];
let queries = [2, 4, 1, 3];
let N = arr.length;
let M = queries.length;
 
findTotNumOfSet(arr, queries, R, N, M);
 
// This code is contributed by saurabh_jaiswal.

输出：

3 2 3 3

时间复杂度： O(M*log R)
辅助空间： O(R)