Q 查询的给定节点的连通分量中的最大元素

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queries[i]包含一个从1到R的整数，每个queries[i]的任务是找到具有值queries[i]的节点的连通分量的最大元素。

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 5 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 maximum element is 3
For the second query: 4 belongs to the set {4, 5} and the maximum element is 5
For the third query: 1 belongs to the set {1, 2, 3} and the maximum element is 3
For the fourth query: 3 belongs to the set {1, 2, 3} and the maximum element is 3

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

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，数组中每个集合的最大值，例如父元素的maxValue[]值。对于每个查询执行查找操作，并为返回的父元素查找maxParent[parent] 。请按照以下步骤解决问题：

• 初始化一个向量maxValue[]以找到每个集合的最大元素。
• 初始化向量parent(R+1), rank(R+1, 0), maxValue(R+1) 。
• 使用变量i遍历范围[1, R+1)并将parent[i]和maxValue[i]的值设置为i 。
• 使用变量i遍历范围[1, N-1]并调用函数操作Union(parent, rank, maxValue, arr[i].first, arr[i].second) 。
• 使用变量i遍历范围[1, M-1]并执行以下步骤：
  • 调用函数操作Find(parent, queries[i]) 。
  • 打印maxValue[i]的值作为结果最大元素。

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
// to find the parent of a disjoint set
int Find(vector& parent, int a)
{
    return parent[a] = (parent[a] == a)
                           ? a
                           : (Find(parent, parent[a]));
}
 
// FUnction to perform union operation
// of disjoint set union
void Union(vector& parent,
           vector& rank,
           vector& maxValue,
           int a, int 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;
    }
 
    parent[b] = a;
 
    // Update the maximum value
    maxValue[a] = max(maxValue[a],
                      maxValue[b]);
}
 
// Function to find the maximum element
// of the set which belongs to the
// element queries[i]
void findMaxValueOfSet(
    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 maxValue of the sets
    vector maxValue(R + 1);
 
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] and
        // maxValue[i] to i
        parent[i] = maxValue[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add arr[i].first and
        // arr[i].second elements to
        // the same set
        Union(parent, rank, maxValue,
              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 maximum value
        // of the set which belongs
        // to the element P
        cout << maxValue[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();
 
    findMaxValueOfSet(arr, queries, R, N, M);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to perform the find operation
// to find the parent of a disjoint set
static int Find(int [] parent, int a)
{
    return parent[a] = (parent[a] == a)
                           ? a
                           : (Find(parent, parent[a]));
}
 
// FUnction to perform union operation
// of disjoint set union
static void Union(int [] parent,
           int [] rank,
           int [] maxValue,
           int a, int 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;
    }
 
    parent[b] = a;
 
    // Update the maximum value
    maxValue[a] = Math.max(maxValue[a],
                      maxValue[b]);
}
 
// Function to find the maximum element
// of the set which belongs to the
// element queries[i]
static void findMaxValueOfSet(
    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 maxValue of the sets
    int [] maxValue = new int[R + 1];
 
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] and
        // maxValue[i] to i
        parent[i] = maxValue[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add arr[i][0] and
        // arr[i][1] elements to
        // the same set
        Union(parent, rank, maxValue,
              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 maximum value
        // of the set which belongs
        // to the element P
        System.out.print(maxValue[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;
 
    findMaxValueOfSet(arr, queries, R, N, M);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python 3 program for the above approach
 
# Function to perform the find operation
# to find the parent of a disjoint set
def Find(parent, a):
    if(parent[parent[a]]!=parent[a]):
        parent[a]=findParent(parent,parent[a])
    return parent[a]
    #return parent[a] = a if (parent[a] == a) else Find(parent, parent[a])
 
# FUnction to perform union operation
# of disjoint set union
def Union(parent, rank, maxValue, a, 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
 
    parent[b] = a
 
    # Update the maximum value
    maxValue[a] = max(maxValue[a],maxValue[b])
 
# Function to find the maximum element
# of the set which belongs to the
# element queries[i]
def findMaxValueOfSet(arr,queries, R, N, M):
    # Stores the parent elements
    # of the sets
    parent = [1 for i in range(R+1)]
 
    # Stores the rank of the sets
    rank = [0 for i in range(R+1)]
 
    # Stores the maxValue of the sets
    maxValue = [0 for i in range(R + 1)]
 
    for i in range(1,R + 1,1):
 
        # Update parent[i] and
        # maxValue[i] to i
        parent[i] = maxValue[i] = i
 
    for i in range(N):
        # Add arr[i].first and
        # arr[i].second elements to
        # the same set
        Union(parent, rank, maxValue, 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 maximum value
        # of the set which belongs
        # to the element P
        print(maxValue[P],end = " ")
 
# Driver Code
if __name__ == '__main__':
    R = 5
    arr = [[1, 2],
           [2, 3],
           [4, 5]];
    queries =  [2, 4, 1, 3]
    N = len(arr)
    M = len(queries)
 
    findMaxValueOfSet(arr, queries, R, N, M)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
 
public class GFG{
 
// Function to perform the find operation
// to find the parent of a disjoint set
static int Find(int [] parent, int a)
{
    return parent[a] = (parent[a] == a)
                           ? a
                           : (Find(parent, parent[a]));
}
 
// FUnction to perform union operation
// of disjoint set union
static void Union(int [] parent,
           int [] rank,
           int [] maxValue,
           int a, int 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;
    }
 
    parent[b] = a;
 
    // Update the maximum value
    maxValue[a] = Math.Max(maxValue[a],
                      maxValue[b]);
}
 
// Function to find the maximum element
// of the set which belongs to the
// element queries[i]
static void findMaxValueOfSet(
    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 maxValue of the sets
    int [] maxValue = new int[R + 1];
 
    for (int i = 1; i < R + 1; i++) {
 
        // Update parent[i] and
        // maxValue[i] to i
        parent[i] = maxValue[i] = i;
    }
 
    for (int i = 0; i < N; i++) {
 
        // Add arr[i,0] and
        // arr[i,1] elements to
        // the same set
        Union(parent, rank, maxValue,
              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 maximum value
        // of the set which belongs
        // to the element P
        Console.Write(maxValue[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.Length;
 
    findMaxValueOfSet(arr, queries, R, N, M);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

// Javascript program for the above approach
 
// Function to perform the find operation
// to find the parent of a disjoint set
function Find(parent, a) {
  return (parent[a] = parent[a] == a ? a : Find(parent, parent[a]));
}
 
// FUnction to perform union operation
// of disjoint set union
function Union(parent, rank, maxValue, a, 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;
  }
 
  parent[b] = a;
 
  // Update the maximum value
  maxValue[a] = Math.max(maxValue[a], maxValue[b]);
}
 
// Function to find the maximum element
// of the set which belongs to the
// element queries[i]
function findMaxValueOfSet(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 maxValue of the sets
  let maxValue = new Array(R + 1);
 
  for (let i = 1; i < R + 1; i++) {
    // Update parent[i] and
    // maxValue[i] to i
    parent[i] = maxValue[i] = i;
  }
 
  for (let i = 0; i < N; i++) {
    // Add arr[i][0] and
    // arr[i][1] elements to
    // the same set
    Union(parent, rank, maxValue, 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 maximum value
    // of the set which belongs
    // to the element P
    document.write(maxValue[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;
 
findMaxValueOfSet(arr, queries, R, N, M);
 
// This code is contributed by ipg2016107

输出：

3 5 3 3

时间复杂度： O(N*log M)
辅助空间： O(N)