给定两个整数arr和P数组，这样在循环之后元素arr [i]将位于arr [P [i]]位置。任务是在数组的所有元素都返回其原始位置后找到最小循环数。
例子：
Input: arr[] = {1, 2, 3}, P[] = {3, 2, 1}
Output: 2
After first move array will be {3, 2, 1}
after second move array will be {1, 2, 3}
Input: arr[] = {4, 5, 1, 2, 3}, P[] = {1, 4, 2, 5, 3}
Output: 4
方法：这个问题似乎是一个典型的数学问题，但是如果我们仔细观察它，我们将发现我们只需要找到置换循环，即连接的分量(由元素运动的循环形成)和每个连接的分量中的节点数将代表该整数在该特定循环中返回其原始位置的时间。
对于整体图，请对每个连接的组件的节点计数进行LCM计算，这就是答案。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to return
// lcm of two numbers
int lcm(int a, int b)
{
return (a * b) / (__gcd(a, b));
}
int dfs(int src, vector adj[], vector &visited)
{
visited[src] = true;
int count = 1;
for (int i = 0; i < adj[src].size(); i++)
if (!visited[adj[src][i]])
count += dfs(adj[src][i], adj, visited);
return count;
}
int findMinTime(int arr[], int P[], int n)
{
// Make a graph
vector adj[n+1];
for (int i = 0; i < n; i++) {
// Add edge
adj[arr[i]].push_back(P[i]);
}
// Count reachable nodes from every node.
vector visited(n+1);
int ans = 1;
for (int i = 0; i < n; i++) {
if (!visited[i]) {
ans = lcm(ans, dfs(i, adj, visited));
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3 };
int P[] = { 3, 2, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMinTime(arr, P, n);
return 0;
} Java
// Java implementation of above approach
import java.util.*;
class GFG
{
// Function to return
// lcm of two numbers
static int lcm(int a, int b)
{
return (a * b) / (__gcd(a, b));
}
static int __gcd(int a, int b)
{
return b == 0 ? a:__gcd(b, a % b);
}
static int dfs(int src, Vector adj[], boolean []visited)
{
visited[src] = true;
int count = 1;
for (int i = 0; i < adj[src].size(); i++)
if (!visited[adj[src].get(i)])
count += dfs(adj[src].get(i), adj, visited);
return count;
}
static int findMinTime(int arr[], int P[], int n)
{
// Make a graph
Vector []adj = new Vector[n + 1];
for (int i = 0; i < n + 1; i++)
adj[i] = new Vector();
for (int i = 0; i < n; i++)
{
// Add edge
adj[arr[i]].add(P[i]);
}
// Count reachable nodes from every node.
boolean []visited = new boolean[n + 1];
int ans = 1;
for (int i = 0; i < n; i++)
{
if (!visited[i])
{
ans = lcm(ans, dfs(i, adj, visited));
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3 };
int P[] = { 3, 2, 1 };
int n = arr.length;
System.out.print(findMinTime(arr, P, n));
}
}
// This code is contributed by Rajput-Ji Python
# Python implementation of above approach
import math
# Function to return
# lcm of two numbers
def lcm(a, b):
return (a * b) // (math.gcd(a, b))
def dfs(src, adj, visited):
visited[src] = True
count = 1
if adj[src] != 0:
for i in range(len(adj[src])):
if (not visited[adj[src][i]]):
count += dfs(adj[src][i], adj, visited)
return count
def findMinTime(arr, P, n):
# Make a graph
adj = [0] * (n + 1)
for i in range(n):
# Add edge
if adj[arr[i]] == 0:
adj[arr[i]] = []
adj[arr[i]].append(P[i])
# Count reachable nodes from every node.
visited = [0] * (n + 1)
ans = 1
for i in range(n):
if (not visited[i]):
ans = lcm(ans, dfs(i, adj, visited))
return ans
# Driver code
arr = [1, 2, 3]
P= [3, 2, 1]
n = len(arr)
print(findMinTime(arr, P, n))
# This code is contributed by shubhamsingh10C#
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return
// lcm of two numbers
static int lcm(int a, int b)
{
return (a * b) / (__gcd(a, b));
}
static int __gcd(int a, int b)
{
return b == 0 ? a:__gcd(b, a % b);
}
static int dfs(int src, List []adj, bool []visited)
{
visited[src] = true;
int count = 1;
for (int i = 0; i < adj[src].Count; i++)
if (!visited[adj[src][i]])
count += dfs(adj[src][i], adj, visited);
return count;
}
static int findMinTime(int []arr, int []P, int n)
{
// Make a graph
List []adj = new List[n + 1];
for (int i = 0; i < n + 1; i++)
adj[i] = new List();
for (int i = 0; i < n; i++)
{
// Add edge
adj[arr[i]].Add(P[i]);
}
// Count reachable nodes from every node.
bool []visited = new bool[n + 1];
int ans = 1;
for (int i = 0; i < n; i++)
{
if (!visited[i])
{
ans = lcm(ans, dfs(i, adj, visited));
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3 };
int []P = { 3, 2, 1 };
int n = arr.Length;
Console.Write(findMinTime(arr, P, n));
}
}
// This code is contributed by 29AjayKumar 输出：
2