索引数组中的组件总数

给定一个包含从0 到 N的N 个整数的数组arr[] ，任务是计算索引数组中组件的数量。

Index array means if we are at ith index then it leads to arr[i]. 
The component of an index array is counted when it forms a cycle. If no cycle persists or the array contains a single element then also we consider it as a component. 
For Example: 
Let array arr[] = {1, 2, 0, 3} 
{1, 2, 0} will form one component as starting from index 0 we reach the index 0 again as: 
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0]) 
 

例子：

Input: arr[] = {1, 2, 3, 5, 0, 4, 6} 
Output: 2 
Explanation: 
Below is the traversal of the 2 components: 
Component 1: Start traversal from 0, then the path of traversal is given by: 
1 -> 2(arr[1]) -> 3(arr[2]) -> 5(arr[3]) -> 4(arr[5]) -> 0(arr[4]) -> 1(arr[0]). 
Component 2: Only 6 is unvisited it creates one more component. 
So, the total components = 2.

Input: arr[] = {1, 2, 0, 3} 
Output: 2 
Explanation: 
Below is the traversal of the 2 components: 
Component 1: Start traversal from 0, then the path of traversal is given by: 
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0]) 
Component 2: Only 3 is unvisited it creates one more component. 
So, the total components = 2. 

做法：思路是利用DFS遍历的概念。以下是步骤：

1. 从第一个未访问的索引开始，该索引将是其中包含整数 0 的索引。
2. 在DFS 遍历期间标记数组中访问过的元素，直到元素形成一个循环。
3. 如果形成一个循环，则意味着我们得到了一个组件，因此增加了组件数量。
4. 对数组中所有未访问的索引重复上述所有步骤，并计算给定索引数组中的总分量。
5. 如果访问了数组的所有索引，则打印连接组件的总数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function for DFS traversal
void dfs(int i, int a[],
         map& m)
{
    // Check if not visited then
    // recurr for the next index
    if (m[i] == 0) {
        m[i] = 1;
 
        // DFS Traversal for next index
        dfs(a[i], a, m);
    }
 
    return;
}
 
// Function for checking which
// indexes are remaining
int allvisited(int a[], int n,
               map& m)
{
    for (int i = 0; i < n; i++) {
 
        // Marks that the ith
        // index is not visited
        if (m[i] == 0)
            return i;
    }
    return -1;
}
 
// Function for counting components
int count(int a[], int n)
{
    int c = 0;
 
    // To mark the visited index
    // during DFS Traversal
    map m;
 
    // Function call
    int x = allvisited(a, n, m);
 
    while (x != -1) {
 
        // Count number of components
        c++;
 
        // DFS call
        dfs(x, a, m);
 
        x = allvisited(a, n, m);
    }
 
    // Print the total count of components
    cout << c << endl;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 4, 3, 5, 0, 2, 6 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    count(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class Main
{
    // Function for DFS traversal
    static void dfs(int i, int[] a,
                    HashMap m)
    {
          
        // Check if not visited then
        // recurr for the next index
        if (!m.containsKey(i))
        {
            m.put(i, 1);
              
            // DFS Traversal for next index
            dfs(a[i], a, m);
        }
        return;
    }
       
    // Function for checking which
    // indexes are remaining
    static int allvisited(int[] a, int n,
                          HashMap m)
    {
        for(int i = 0; i < n; i++)
        {
              
            // Marks that the ith
            // index is not visited
            if (!m.containsKey(i))
                return i;
        }
        return -1;
    }
       
    // Function for counting components
    static void count(int[] a, int n)
    {
        int c = 0;
          
        // To mark the visited index
        // during DFS Traversal
        HashMap m = new HashMap<>();
                                             
        // Function call
        int x = allvisited(a, n, m);
       
        while (x != -1)
        {
              
            // Count number of components
            c++;
              
            // DFS call
            dfs(x, a, m);
       
            x = allvisited(a, n, m);
        }
          
        // Print the total count of components
        System.out.print(c);
    }
 
    public static void main(String[] args)
    {
       
        // Given array arr[]
        int[] arr = { 1, 4, 3, 5, 0, 2, 6 };
       
        int N = arr.length;
       
        // Function Call
        count(arr, N);
    }
}
 
// This code is contributed by divyesh072019

Python3

# Python3 program for the above approach
  
# Function for DFS traversal
def dfs(i, a, m):
 
    # Check if not visited then
    # recurr for the next index
    if i in m:
        if m[i] == 0:
            m[i] = 1
             
            # DFS Traversal for next index
            dfs(a[i], a, m)
    else:
        m[i] = 1
         
        # DFS Traversal for next index
        dfs(a[i], a, m)
  
    return
 
# Function for checking which
# indexes are remaining
def allvisited(a, n, m):
     
    for i in range(n):
  
        # Marks that the ith
        # index is not visited
        if i in m:
            if m[i] == 0:
                return i
         
        else:
            return i
 
    return -1
 
# Function for counting components
def count(a, n):
 
    c = 0
  
    # To mark the visited index
    # during DFS Traversal
    m = dict()
  
    # Function call
    x = allvisited(a, n, m)
  
    while (x != -1):
  
        # Count number of components
        c += 1
  
        # DFS call
        dfs(x, a, m)
  
        x = allvisited(a, n, m)
     
    # Print the total count of components
    print(c)
     
# Driver Code
if __name__=='__main__':
 
    # Given array arr[]
    arr = [ 1, 4, 3, 5, 0, 2, 6 ]
     
    N = len(arr)
  
    # Function Call
    count(arr, N)
  
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
  
// Function for DFS traversal
static void dfs(int i, int []a,
     Dictionary m)
{
     
    // Check if not visited then
    // recurr for the next index
    if (!m.ContainsKey(i))
    {
        m[i] = 1;
         
        // DFS Traversal for next index
        dfs(a[i], a, m);
    }
    return;
}
  
// Function for checking which
// indexes are remaining
static int allvisited(int []a, int n,
               Dictionary m)
{
    for(int i = 0; i < n; i++)
    {
         
        // Marks that the ith
        // index is not visited
        if (!m.ContainsKey(i))
            return i;
    }
    return -1;
}
  
// Function for counting components
static void count(int []a, int n)
{
    int c = 0;
     
    // To mark the visited index
    // during DFS Traversal
    Dictionary m = new Dictionary();
                                        
    // Function call
    int x = allvisited(a, n, m);
  
    while (x != -1)
    {
         
        // Count number of components
        c++;
         
        // DFS call
        dfs(x, a, m);
  
        x = allvisited(a, n, m);
    }
     
    // Print the total count of components
    Console.Write(c);
}
  
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = { 1, 4, 3, 5, 0, 2, 6 };
  
    int N = arr.Length;
  
    // Function Call
    count(arr, N);
}
}
 
// This code is contributed by pratham76

Javascript

 
// Javascript program for the above approach
 
// Function for DFS traversal
function dfs(i, a, m)
{
    // Check if not visited then
    // recurr for the next index
    if (!m.has(i)) {
        m.set(i, 1);
        m[i] = 1;
 
        // DFS Traversal for next index
        dfs(a[i], a, m);
    }
 
    return;
}
 
// Function for checking which
// indexes are remaining
function allvisited(a, n, m)
{
    for (var i = 0; i < n; i++) {
 
        // Marks that the ith
        // index is not visited
        if (!m.has(i))
            return i;
    }
    return -1;
}
 
// Function for counting components
function count(a, n)
{
    var c = 0;
 
    // To mark the visited index
    // during DFS Traversal
    var m = new Map();
 
    // Function call
    var x = allvisited(a, n, m);
 
    while (x != -1) {
 
        // Count number of components
        c++;
 
        // DFS call
        dfs(x, a, m);
 
        x = allvisited(a, n, m);
    }
 
    // Print the total count of components
    document.write( c );
}
 
// Driver Code
 
// Given array arr[]
var arr = [1, 4, 3, 5, 0, 2, 6];
var N = arr.length;
 
// Function Call
count(arr, N);
 
// This code is contributed by itsok.

输出：

3

时间复杂度： O(N)
辅助空间： O(N)

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