📌  相关文章
📜  构造一棵树,该树的所有根到叶路径的节点总数不能被该路径中的节点数整除

📅  最后修改于: 2021-04-17 12:58:44             🧑  作者: Mango

鉴于由N N叉树节点编号从1到N节点1根,任务是分配值给树的每个节点,使得值的总和从任何根,其含有至少两个叶路径节点不能被该路径上的节点数整除。

例子:

方法:给定的问题可以基于这样的观察,对于任何根到叶路径以2号节点至少,如果说k为解决值沿着这条道路的谎言K2 * K独家之间的总和那么和可以永远不会被K整除,因为(K,2 * K)范围内的任何数字都不会被K整除。因此,对于K = 1 ,将奇数级节点的节点值分配为1 ,其余部分分配为2 。请按照以下步骤解决问题:

  • 初始化一个数组,说出大小为N + 1的answer []以存储分配给节点的值。
  • 将变量K初始化为1,以将值分配给每个节点。
  • 初始化用于在给定树上执行BFS遍历的队列,并将节点中的值为1的节点推送到队列中,并将该节点的值初始化为1
  • 迭代直到队列为非空,然后执行以下步骤:
    • 弹出队列的前端节点,如果分配给弹出节点的值为1,则将K的值更新为2 。否则,将K更新为1
    • 遍历当前弹出节点的所有子节点,并将该子节点推送到队列中,并将值K分配给该子节点。
  • 完成上述步骤后,打印存储在数组answer []中的值作为结果。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function to assign values to nodes
// of the tree s.t. sum of values of
// nodes of path between any 2 nodes
// is not divisible by length of path
void assignValues(int Edges[][2], int n)
{
    // Stores the adjacency list
    vector  tree[n + 1];
     
      // Create a adjacency list
      for(int i = 0; i < n - 1; i++) {
       
      int u = Edges[i][0];
      int v = Edges[i][1];
      tree[u].push_back(v);
      tree[v].push_back(u);
    }
   
    // Stores whether node is
      // visited or not
    vector  visited(n + 1, false);
 
    // Stores the node values
    vector  answer(n + 1);
 
    // Variable used to assign values to
      // the nodes alternatively to the
      // parent child
    int K = 1;
 
    // Declare a queue
    queue  q;
 
    // Push the 1st node
    q.push(1);
 
    // Assign K value to this node
    answer[1] = K;
 
    while (!q.empty()) {
 
        // Dequeue the node
        int node = q.front();
        q.pop();
 
        // Mark it as visited
        visited[node] = true;
 
        // Upgrade the value of K
        K = ((answer[node] == 1) ? 2 : 1);
 
        // Assign K to the child nodes
        for (auto child : tree[node]) {
 
            // If the child is unvisited
            if (!visited[child]) {
 
                // Enqueue the child
                q.push(child);
 
                // Assign K to the child
                answer[child] = K;
            }
        }
    }
     
      // Print the value assigned to
      // the nodes
    for (int i = 1; i <= n; i++) {
        cout << answer[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 11;
    int Edges[][2] = {{1, 2}, {1, 3}, {1, 4},
                      {1, 5}, {2, 6}, {2, 10},
                      {10, 11}, {3, 7}, {4, 8},
                      {5, 9}};
 
    // Function Call
    assignValues(Edges, N);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to assign values to nodes
// of the tree s.t. sum of values of
// nodes of path between any 2 nodes
// is not divisible by length of path
static void assignValues(int Edges[][], int n)
{
     
    // Stores the adjacency list
    ArrayList> tree = new ArrayList<>();
 
    for(int i = 0; i < n + 1; i++)
        tree.add(new ArrayList<>());
 
    // Create a adjacency list
    for(int i = 0; i < n - 1; i++)
    {
        int u = Edges[i][0];
        int v = Edges[i][1];
        tree.get(u).add(v);
        tree.get(v).add(u);
    }
 
    // Stores whether node is
    // visited or not
    boolean[] visited = new boolean[n + 1];
 
    // Stores the node values
    int[] answer = new int[n + 1];
 
    // Variable used to assign values to
    // the nodes alternatively to the
    // parent child
    int K = 1;
 
    // Declare a queue
    Queue q = new LinkedList<>();
 
    // Push the 1st node
    q.add(1);
 
    // Assign K value to this node
    answer[1] = K;
 
    while (!q.isEmpty())
    {
 
        // Dequeue the node
        int node = q.peek();
        q.poll();
 
        // Mark it as visited
        visited[node] = true;
 
        // Upgrade the value of K
        K = ((answer[node] == 1) ? 2 : 1);
 
        // Assign K to the child nodes
        for(Integer child : tree.get(node))
        {
 
            // If the child is unvisited
            if (!visited[child])
            {
                 
                // Enqueue the child
                q.add(child);
 
                // Assign K to the child
                answer[child] = K;
            }
        }
    }
 
    // Print the value assigned to
    // the nodes
    for(int i = 1; i <= n; i++)
    {
        System.out.print(answer[i] + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 11;
    int Edges[][] = { { 1, 2 }, { 1, 3 }, 
                      { 1, 4 }, { 1, 5 },
                      { 2, 6 }, { 2, 10 },
                      { 10, 11 }, { 3, 7 },
                      { 4, 8 }, { 5, 9 } };
 
    // Function Call
    assignValues(Edges, N);
}
}
 
// This code is contributed by offbeat


Python3
# Python3 program for the above approach
from collections import deque
 
# Function to assign values to nodes
# of the tree s.t. sum of values of
# nodes of path between any 2 nodes
# is not divisible by length of path
def assignValues(Edges, n):
   
    # Stores the adjacency list
    tree = [[] for i in range(n + 1)]
 
    # Create a adjacency list
    for i in range(n - 1):
 
        u = Edges[i][0]
        v = Edges[i][1]
        tree[u].append(v)
        tree[v].append(u)
 
    # Stores whether any node is
    # visited or not
    visited = [False]*(n+1)
 
    # Stores the node values
    answer = [0]*(n + 1)
 
    # Variable used to assign values to
    # the nodes alternatively to the
    # parent child
    K = 1
 
    # Declare a queue
    q = deque()
 
    # Push the 1st node
    q.append(1)
 
    # Assign K value to this node
    answer[1] = K
 
    while (len(q) > 0):
 
        # Dequeue the node
        node = q.popleft()
        # q.pop()
 
        # Mark it as visited
        visited[node] = True
 
        # Upgrade the value of K
        K = 2 if (answer[node] == 1) else 1
 
        # Assign K to the child nodes
        for child in tree[node]:
 
            # If the child is unvisited
            if (not visited[child]):
 
                # Enqueue the child
                q.append(child)
 
                # Assign K to the child
                answer[child] = K
 
    # Prthe value assigned to
    # the nodes
    for i in range(1, n + 1):
        print(answer[i],end=" ")
 
# Driver Code
if __name__ == '__main__':
    N = 7
    Edges = [ [ 1, 2 ], [ 4, 6 ],
               [ 3, 5 ], [ 1, 4 ],
               [ 7, 5 ], [ 5, 1 ] ]
 
    # Function Call
    assignValues(Edges, N)
 
# This code is contributed by mohit kumar 29.


输出:
1 2 2 2 2 1 1 1 1 1 2

时间复杂度: O(V + E)
辅助空间: O(V + E)