根据给定条件计算最多N个可以形成非循环图的整数的排列

给定整数N，任务是根据以下条件从可以形成无环图的范围[1，N]中查找整数的排列数目：

• 对于每1≤i≤N ，找到最大j ，使1≤j 且A [j]> A [i ]，并在节点i和节点j之间添加无向边。
• 对于每1≤i≤N ，找到最小的j ，使i 且A [j]> A [i] ，并在节点i和节点j之间添加无向边。

例子：

Input: 3
Output: 4
Explanation: 
{1, 2, 3}: Possible (Graph Edges : { [1, 2], [2, 3] }. Therefore, no cycle exists) 
{1, 3, 2}: Possible (Graph Edges : { [1, 3], [2, 3] }. Therefore, no cycle exists) 
{2, 1, 3}: Not possible (Graph Edges : { [2, 3], [1, 3], [1, 2] }. Therefore, cycle exists) 
{2, 3, 1}: Possible (Graph Edges : { [2, 3], [1, 3] }. Therefore, no cycle exists) 
{3, 1, 2}: Not possible (Graph Edges : { [1, 2], [1, 3], [2, 3] }. Therefore, cycle exists) 
{3, 2, 1}: Possible (Graph Edges : { [2, 3], [1, 2] }. Therefore, no cycle exists)

Input : 4
Output : 8

方法：请按照以下步骤解决问题：

• 共有N个！可能生成的可能数组，包含范围为[1，N]的整数。
• 根据给定的条件，如果i，j，k(i 是数组的索引，则如果A [i]> A [j]和A [j] ，则将存在是由边{ [A [j]，A [i]]，[A [j]，A [k]]，[A [i]，A [k]]}组成的循环。
• 除去这些置换，剩余2 ^{(N – 1)个}可能的置换。
• 对于[1，N – 1]范围内的整数，剩余的排列仅给出两个可能的位置，而对于N则给出1个可能的位置。

Illustration: 
For N = 3: 
Permutation {2, 1, 3} contains a cycle as A[0] > A[1] and A[1] < A[2]. 
Permutation {3, 1, 2} contains a cycle as A[0] > A[1] and A[1] < A[2].
All remaining permutations can generate an acyclic graph. 
Therefore, the count of valid permutations = 3! – 2 = 4 = 2^{3 – 1} 

• 因此，打印2 ^{N – 1}作为要求的答案。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Find the count of possible graphs
void possibleAcyclicGraph(int N)
{
    cout << pow(2, N - 1);
    return;
}
 
// Driver Code
int main()
{
 
    int N = 4;
    possibleAcyclicGraph(N);
 
    return 0;
}

Java

// Java implementation of
// the above approach
import java.util.*;
class GFG{
 
// Find the count of
// possible graphs
static void possibleAcyclicGraph(int N)
{
  System.out.print((int)Math.pow(2,
                                 N - 1));
  return;
}
 
// Driver Code
public static void main(String[] args)
{
  int N = 4;
  possibleAcyclicGraph(N);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of above approach
 
# Find the count of possible graphs
def possibleAcyclicGraph(N):
     
    print(pow(2, N - 1))
    return
 
# Driver code
if __name__ == '__main__':
     
    N = 4
     
    possibleAcyclicGraph(N)
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of
// the above approach
using System;
 
class GFG{
     
// Find the count of
// possible graphs
static void possibleAcyclicGraph(int N)
{
    Console.Write((int)Math.Pow(2, N - 1));
     
    return;
}
  
// Driver Code
public static void Main()
{
    int N = 4;
     
    possibleAcyclicGraph(N);
}
}
 
// This code is contributed by sanjoy_62

输出

8

时间复杂度：O(logN)
空间复杂度：O(1)