📜  使二进制搜索树

📅  最后修改于: 2021-06-26 09:00:34             🧑  作者: Mango

给定大小为N的数组arr [] 。任务是查找是否有可能用给定的元素数组制作二叉搜索树,以使通过公共边连接的任意两个顶点的最大公共除数> 1 。如果可能,请打印“是”,否则打印“否”

例子:

方法:DP(l,r,root)为DP,以确定是否有可能从子段[l..r]组装以根为根的树。
很容易看出,计算它需要提取[l..root – 1]留下了这样的根和根直接从[根+ 1..right]这样的:

  • gcd()> 1
  • gcd(一个root ,一个root)> 1
  • DP(l,root-1,root left )= 1
  • DP(根+1,r,根)= 1

只要给定[l..r]的所有子段的所有DP(x,y,z)值,就可以在O(r – l)中完成。考虑总共O(n 3 )DP状态,最终的复杂度为O(n 4 ),这太多了。

让我们将DP转换成状态为0或1的DPnew(l,r,state)。立即发现DP(l,r,root)继承自DPnew(l,root-1,1)和DPnew(根+1,r,0)。现在我们具有O(n 2 )状态,但是同时,所有转换都在线性时间内执行。因此,最终复杂度为O(n 3 ),足以通过。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Maxium number of vertices
#define N 705
  
// To store is it possible at
// particular pace or not
int dp[N][N][2];
  
// Return 1 if from l to r, it is possible with
// the given state
int possibleWithState(int l, int r, int state, int a[])
{
    // Base condition
    if (l > r)
        return 1;
  
    // If it is already calculated
    if (dp[l][r][state] != -1)
        return dp[l][r][state];
  
    // Choose the root
    int root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
  
    // Traverse in range l to r
    for (int i = l; i <= r; i++) {
  
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1) {
            int x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
            int y = possibleWithState(i + 1, r, 0, a);
            if (x == 1 && y == 1)
                return dp[l][r][state] = 1;
        }
    }
  
    // If not possible
    return dp[l][r][state] = 0;
}
  
// Function that return true if it is possible
// to make Binary Search Tree
bool isPossible(int a[], int n)
{
    memset(dp, -1, sizeof dp);
  
    // Sort the given array
    sort(a, a + n);
  
    // Check it is possible rooted at i
    for (int i = 0; i < n; i++)
  
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a)
            && possibleWithState(i + 1, n - 1, 0, a)) {
            return true;
        }
  
    return false;
}
  
// Driver code
int main()
{
    int a[] = { 3, 6, 9, 18, 36, 108 };
    int n = sizeof(a) / sizeof(a[0]);
  
    if (isPossible(a, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}


Java
// Java implementation of the approach
import java.util.*;
class GFG
{
      
static int __gcd(int a, int b) 
{ 
      
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
      
    // base case 
    if (a == b) 
        return a; 
      
    // a is greater 
    if (a > b) 
        return __gcd(a - b, b); 
    return __gcd(a, b-a); 
} 
  
// Maxium number of vertices
static final int N = 705;
  
// To store is it possible at
// particular pace or not
static int dp[][][] = new int[N][N][2];
  
// Return 1 if from l to r, it is 
// possible with the given state
static int possibleWithState(int l, int r,
                        int state, int a[])
{
    // Base condition
    if (l > r)
        return 1;
  
    // If it is already calculated
    if (dp[l][r][state] != -1)
        return dp[l][r][state];
  
    // Choose the root
    int root;
    if (state == 1)
        root = a[r + 1];
    else
        root = a[l - 1];
  
    // Traverse in range l to r
    for (int i = l; i <= r; i++) 
    {
  
        // If gcd is greater than one
        // check for both sides
        if (__gcd(a[i], root) > 1) 
        {
            int x = possibleWithState(l, i - 1, 1, a);
            if (x != 1)
                continue;
                  
            int y = possibleWithState(i + 1, r, 0, a);
              
            if (x == 1 && y == 1)
                return dp[l][r][state] = 1;
        }
    }
  
    // If not possible
    return dp[l][r][state] = 0;
}
  
// Function that return true if it is possible
// to make Binary Search Tree
static boolean isPossible(int a[], int n)
{
    for(int i = 0; i < dp.length; i++)
        for(int j = 0; j < dp[i].length; j++)
            for(int k = 0; k < dp[i][j].length; k++)
                dp[i][j][k]=-1;
  
    // Sort the given array
    Arrays.sort(a);
  
    // Check it is possible rooted at i
    for (int i = 0; i < n; i++)
  
        // Check at both sides
        if (possibleWithState(0, i - 1, 1, a) != 0 && 
            possibleWithState(i + 1, n - 1, 0, a) != 0)
        {
            return true;
        }
  
    return false;
}
  
// Driver code
public static void main(String args[])
{
    int a[] = { 3, 6, 9, 18, 36, 108 };
    int n = a.length;
  
    if (isPossible(a, n))
        System.out.println("Yes");
    else
        System.out.println("No");
  
}
}
  
// This code is contributed by 
// Arnab Kundu


Python3
# Python3 implementation of the approach 
import math
  
# Maxium number of vertices 
N = 705
  
# To store is it possible at 
# particular pace or not 
dp = [[[-1 for z in range(2)] 
           for x in range(N)] 
           for y in range(N)]
  
# Return 1 if from l to r, it is 
# possible with the given state 
def possibleWithState(l, r, state, a): 
  
    # Base condition 
    if (l > r):
        return 1
  
    # If it is already calculated 
    if (dp[l][r][state] != -1):
        return dp[l][r][state] 
  
    # Choose the root 
    root = 0
    if (state == 1) :
        root = a[r + 1] 
    else:
        root = a[l - 1] 
  
    # Traverse in range l to r 
    for i in range(l, r + 1): 
  
        # If gcd is greater than one 
        # check for both sides 
        if (math.gcd(a[i], root) > 1): 
            x = possibleWithState(l, i - 1, 1, a) 
            if (x != 1): 
                continue
            y = possibleWithState(i + 1, r, 0, a) 
            if (x == 1 and y == 1) :
                return 1
  
    # If not possible 
    return 0
  
# Function that return true if it is 
# possible to make Binary Search Tree 
def isPossible(a, n): 
      
    # Sort the given array 
    a.sort() 
  
    # Check it is possible rooted at i 
    for i in range(n):
          
        # Check at both sides 
        if (possibleWithState(0, i - 1, 1, a) and 
            possibleWithState(i + 1, n - 1, 0, a)):
            return True
              
    return False
  
# Driver Code 
if __name__ == '__main__': 
    a = [3, 6, 9, 18, 36, 108]
    n = len(a) 
    if (isPossible(a, n)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#
// C# implementation of the approach 
using System; 
  
class GFG 
{ 
      
static int __gcd(int a, int b) 
{ 
      
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
      
    // base case 
    if (a == b) 
        return a; 
      
    // a is greater 
    if (a > b) 
        return __gcd(a - b, b); 
    return __gcd(a, b-a); 
} 
  
// Maximum number of vertices 
static int N = 705; 
  
// To store is it possible at 
// particular pace or not 
static int [,,]dp = new int[N, N, 2]; 
  
// Return 1 if from l to r, it is 
// possible with the given state 
static int possibleWithState(int l, int r, 
                        int state, int []a) 
{ 
    // Base condition 
    if (l > r) 
        return 1; 
  
    // If it is already calculated 
    if (dp[l, r, state] != -1) 
        return dp[l, r, state]; 
  
    // Choose the root 
    int root; 
    if (state == 1) 
        root = a[r + 1]; 
    else
        root = a[l - 1]; 
  
    // Traverse in range l to r 
    for (int i = l; i <= r; i++) 
    { 
  
        // If gcd is greater than one 
        // check for both sides 
        if (__gcd(a[i], root) > 1) 
        { 
            int x = possibleWithState(l, i - 1, 1, a); 
            if (x != 1) 
                continue; 
                  
            int y = possibleWithState(i + 1, r, 0, a); 
              
            if (x == 1 && y == 1) 
                return dp[l,r,state] = 1; 
        } 
    } 
  
    // If not possible 
    return dp[l,r,state] = 0; 
} 
  
// Function that return true 
// if it is possible to make 
// Binary Search Tree 
static bool isPossible(int []a, int n) 
{ 
    for(int i = 0; i < dp.GetLength(0); i++) 
        for(int j = 0; j < dp.GetLength(1); j++) 
            for(int k = 0; k < dp.GetLength(2); k++) 
                dp[i, j, k]=-1; 
  
    // Sort the given array 
    Array.Sort(a); 
  
    // Check it is possible rooted at i 
    for (int i = 0; i < n; i++) 
  
        // Check at both sides 
        if (possibleWithState(0, i - 1, 1, a) != 0 && 
            possibleWithState(i + 1, n - 1, 0, a) != 0) 
        { 
            return true; 
        } 
  
    return false; 
} 
  
// Driver code 
public static void Main(String []args) 
{ 
    int []a = { 3, 6, 9, 18, 36, 108 }; 
    int n = a.Length; 
  
    if (isPossible(a, n)) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
  
} 
} 
  
// This code is contributed by 29AjayKumar


输出:
Yes

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