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📜  无向图的连通分量中所有节点的最大 GCD

📅  最后修改于: 2021-09-04 13:05:13             🧑  作者: Mango

给定一个由V个顶点和一个表示节点对之间边的二维数组E[][2]组成的无向图。给定另一个数组arr[]表示分配给每个节点的值,任务是在图中所有连接组件的 GCD 中找到最大的 GCD。

例子:

方法:可以通过对给定图执行深度优先搜索遍历,然后在所有连通分量中找到最大 GCD 来解决给定问题。请按照以下步骤解决问题:

  • 初始化一个变量,比如maxGCDINT_MIN ,以存储所有连接组件之间的最大 GCD。
  • 初始化另一个变量,比如currentGCD0 ,以独立存储每个连接组件的 GCD。
  • 将一个辅助数组visited[]初始化为false,以在DFS Traversal 中存储访问过的节点。
  • 在范围[1, V] 上迭代每个顶点并执行以下步骤:
    • 如果当前顶点未被访问,即visited[i] = false ,则将currentGCD初始化为0
    • 从与currentGCD值和更新currentGCD值作为currentGCD的GCD当前顶点执行DFS遍历和常用3 [I – 1]中的每个递归调用。
    • 如果currentGCD的值大于maxGCD ,则将maxGCD更新为currentGCD
  • 完成以上步骤后,打印maxGCD的值作为结果。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the  GCD of two
// numbers a and b
int gcd(int a, int b)
{
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    return gcd(b, a % b);
}
 
// Function to perform DFS Traversal
void depthFirst(int v, vector graph[],
                vector& visited,
                int& currGCD,
                vector values)
{
    // Mark the visited vertex as true
    visited[v] = true;
 
    // Update GCD of current
    // connected component
    currGCD = gcd(currGCD, values[v - 1]);
 
    // Traverse all adjacent nodes
    for (auto child : graph[v]) {
 
        if (visited[child] == false) {
 
            // Recursive call to perform
            // DFS traversal
            depthFirst(child, graph, visited,
                       currGCD, values);
        }
    }
}
 
// Function to find the maximum GCD
// of nodes among all the connected
// components of an undirected graph
void maximumGcd(int Edges[][2], int E,
                int V, vector& arr)
{
    vector graph[V + 1];
 
    // Traverse the edges
    for (int i = 0; i < E; i++) {
 
        int u = Edges[i][0];
        int v = Edges[i][1];
 
        graph[u].push_back(v);
        graph[v].push_back(u);
    }
 
    // Initialize boolean array
    // to mark visited vertices
    vector visited(V + 1, false);
 
    // Stores the maximum GCD value
    int maxGCD = INT_MIN;
 
    // Traverse all the vertices
    for (int i = 1; i <= V; i++) {
 
        // If node is not visited
        if (visited[i] == false) {
 
            // Stores GCD of current
            // connected component
            int currGCD = 0;
 
            // Perform DFS Traversal
            depthFirst(i, graph, visited,
                       currGCD, arr);
 
            // Update maxGCD
            if (currGCD > maxGCD) {
                maxGCD = currGCD;
            }
        }
    }
 
    // Print the result
    cout << maxGCD;
}
 
// Driver Code
int main()
{
    int E = 3, V = 5;
    vector arr = { 23, 43, 123, 54, 2 };
    int Edges[][2] = { { 1, 3 }, { 2, 3 }, { 1, 2 } };
 
    maximumGcd(Edges, E, V, arr);
 
    return 0;
}


Java
// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
  static int currGCD;
 
  // Function to find the  GCD of two
  // numbers a and b
  static int gcd(int a, int b)
  {
    // Base Case
    if (b == 0)
      return a;
 
    // Recursively find the GCD
    return gcd(b, a % b);
  }
 
  // Function to perform DFS Traversal
  static void depthFirst(int v,
                         ArrayList graph[],
                         boolean visited[], int values[])
  {
    // Mark the visited vertex as true
    visited[v] = true;
 
    // Update GCD of current
    // connected component
    currGCD = gcd(currGCD, values[v - 1]);
 
    // Traverse all adjacent nodes
    for (int child : graph[v]) {
 
      if (visited[child] == false) {
 
        // Recursive call to perform
        // DFS traversal
        depthFirst(child, graph, visited, values);
      }
    }
  }
 
  // Function to find the maximum GCD
  // of nodes among all the connected
  // components of an undirected graph
  static void maximumGcd(int Edges[][], int E, int V,
                         int arr[])
  {
 
    ArrayList graph[] = new ArrayList[V + 1];
 
    // Initialize the graph
    for (int i = 0; i < V + 1; i++)
      graph[i] = new ArrayList<>();
 
    // Traverse the edges
    for (int i = 0; i < E; i++) {
 
      int u = Edges[i][0];
      int v = Edges[i][1];
 
      graph[u].add(v);
      graph[v].add(u);
    }
 
    // Initialize boolean array
    // to mark visited vertices
    boolean visited[] = new boolean[V + 1];
 
    // Stores the maximum GCD value
    int maxGCD = Integer.MIN_VALUE;
 
    // Traverse all the vertices
    for (int i = 1; i <= V; i++) {
 
      // If node is not visited
      if (visited[i] == false) {
 
        // Stores GCD of current
        // connected component
        currGCD = 0;
 
        // Perform DFS Traversal
        depthFirst(i, graph, visited, arr);
 
        // Update maxGCD
        if (currGCD > maxGCD) {
          maxGCD = currGCD;
        }
      }
    }
 
    // Print the result
    System.out.println(maxGCD);
  }
   
  // Driver Code
  public static void main(String[] args)
  {
 
    int E = 3, V = 5;
    int arr[] = { 23, 43, 123, 54, 2 };
    int Edges[][] = { { 1, 3 }, { 2, 3 }, { 1, 2 } };
 
    maximumGcd(Edges, E, V, arr);
  }
}
 
// Thi code is contributed by Kingash.


Python3
# Python 3 program for the above approach
from math import gcd
import sys
 
# Function to find the  GCD of two
# numbers a and b
currGCD = 0
 
# Function to perform DFS Traversal
def depthFirst(v, graph, visited, values):
    global currGCD
     
    # Mark the visited vertex as true
    visited[v] = True
 
    # Update GCD of current
    # connected component
    currGCD = gcd(currGCD, values[v - 1])
 
    # Traverse all adjacent nodes
    for child in graph[v]:
        if (visited[child] == False):
           
            # Recursive call to perform
            # DFS traversal
            depthFirst(child, graph, visited, values)
 
# Function to find the maximum GCD
# of nodes among all the connected
# components of an undirected graph
def maximumGcd(Edges, E, V, arr):
    global currGCD
    graph = [[] for i in range(V + 1)]
 
    # Traverse the edges
    for i in range(E):
        u = Edges[i][0]
        v = Edges[i][1]
 
        graph[u].append(v)
        graph[v].append(u)
 
    # Initialize boolean array
    # to mark visited vertices
    visited = [False for i in range(V+1)]
 
    # Stores the maximum GCD value
    maxGCD = -sys.maxsize - 1
 
    # Traverse all the vertices
    for i in range(1, V + 1, 1):
       
        # If node is not visited
        if (visited[i] == False):
           
            # Stores GCD of current
            # connected component
            currGCD = 0
 
            # Perform DFS Traversal
            depthFirst(i, graph, visited, arr)
 
            # Update maxGCD
            if (currGCD > maxGCD):
                maxGCD = currGCD
 
    # Print the result
    print(maxGCD)
 
# Driver Code
if __name__ == '__main__':
    E = 3
    V = 5
    arr =  [23, 43, 123, 54, 2]
    Edges =  [[1, 3 ], [2, 3], [1, 2]]
    maximumGcd(Edges, E, V, arr)
 
    # This code is contributed by ipg2016107.


输出:
54

时间复杂度: O((V + E) * log(M)),其中M给定数组arr[]的最小元素
辅助空间: O(V)

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