给定一个有N个顶点和M 个边的无向图,任务是打印给定图中度为素数的所有节点。
例子:
Input: N = 4, arr[][] = { { 1, 2 }, { 1, 3 }, { 1, 4 }, { 2, 3 }, { 2, 4 }, { 3, 4 } }
Output: 1 2 3 4
Explanation:
Below is the graph for the above information:
The degree of the node as per above graph is:
Node -> Degree
1 -> 3
2 -> 3
3 -> 3
4 -> 3
Hence, the nodes with prime degree are 1 2 3 4
Input: N = 5, arr[][] = { { 1, 2 }, { 1, 3 }, { 2, 4 }, { 2, 5 } }
Output: 1
方法:
- 使用埃拉托色尼筛法计算最大为 10 5的素数。
- 对于每个顶点,度数可以通过给定图在相应顶点的邻接表的长度来计算。
- 打印给定图中度数为素数的顶点。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
int n = 10005;
// To store Prime Numbers
vector Prime(n + 1, true);
// Function to find the prime numbers
// till 10^5
void SieveOfEratosthenes()
{
int i, j;
Prime[0] = Prime[1] = false;
for (i = 2; i * i <= 10005; i++) {
// Traverse all multiple of i
// and make it false
if (Prime[i]) {
for (j = 2 * i; j < 10005; j += i) {
Prime[j] = false;
}
}
}
}
// Function to print the nodes having
// prime degree
void primeDegreeNodes(int N, int M,
int edges[][2])
{
// To store Adjacency List of
// a Graph
vector Adj[N + 1];
// Make Adjacency List
for (int i = 0; i < M; i++) {
int x = edges[i][0];
int y = edges[i][1];
Adj[x].push_back(y);
Adj[y].push_back(x);
}
// To precompute prime numbers
// till 10^5
SieveOfEratosthenes();
// Traverse each vertex
for (int i = 1; i <= N; i++) {
// Find size of Adjacency List
int x = Adj[i].size();
// If length of Adj[i] is Prime
// then print it
if (Prime[x])
cout << i << ' ';
}
}
// Driver code
int main()
{
// Vertices and Edges
int N = 4, M = 6;
// Edges
int edges[M][2] = { { 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 } };
// Function Call
primeDegreeNodes(N, M, edges);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG{
static int n = 10005;
// To store Prime Numbers
static boolean []Prime = new boolean[n + 1];
// Function to find the prime numbers
// till 10^5
static void SieveOfEratosthenes()
{
int i, j;
Prime[0] = Prime[1] = false;
for (i = 2; i * i <= 10005; i++)
{
// Traverse all multiple of i
// and make it false
if (Prime[i])
{
for (j = 2 * i; j < 10005; j += i)
{
Prime[j] = false;
}
}
}
}
// Function to print the nodes having
// prime degree
static void primeDegreeNodes(int N, int M,
int edges[][])
{
// To store Adjacency List of
// a Graph
Vector []Adj = new Vector[N + 1];
for(int i = 0; i < Adj.length; i++)
Adj[i] = new Vector();
// Make Adjacency List
for (int i = 0; i < M; i++)
{
int x = edges[i][0];
int y = edges[i][1];
Adj[x].add(y);
Adj[y].add(x);
}
// To precompute prime numbers
// till 10^5
SieveOfEratosthenes();
// Traverse each vertex
for (int i = 1; i <= N; i++)
{
// Find size of Adjacency List
int x = Adj[i].size();
// If length of Adj[i] is Prime
// then print it
if (Prime[x])
System.out.print(i + " ");
}
}
// Driver code
public static void main(String[] args)
{
// Vertices and Edges
int N = 4, M = 6;
// Edges
int edges[][] = { { 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 } };
Arrays.fill(Prime, true);
// Function Call
primeDegreeNodes(N, M, edges);
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 implementation of
# the above approach
n = 10005;
# To store Prime Numbers
Prime = [True for i in range(n + 1)]
# Function to find
# the prime numbers
# till 10^5
def SieveOfEratosthenes():
i = 2
Prime[0] = Prime[1] = False;
while i * i <= 10005:
# Traverse all multiple
# of i and make it false
if (Prime[i]):
for j in range(2 * i, 10005):
Prime[j] = False
i += 1
# Function to print the
# nodes having prime degree
def primeDegreeNodes(N, M, edges):
# To store Adjacency
# List of a Graph
Adj = [[] for i in range(N + 1)];
# Make Adjacency List
for i in range(M):
x = edges[i][0];
y = edges[i][1];
Adj[x].append(y);
Adj[y].append(x);
# To precompute prime
# numbers till 10^5
SieveOfEratosthenes();
# Traverse each vertex
for i in range(N + 1):
# Find size of Adjacency List
x = len(Adj[i]);
# If length of Adj[i] is Prime
# then print it
if (Prime[x]):
print(i, end = ' ')
# Driver code
if __name__ == "__main__":
# Vertices and Edges
N = 4
M = 6
# Edges
edges = [[1, 2], [1, 3],
[1, 4], [2, 3],
[2, 4], [3, 4]];
# Function Call
primeDegreeNodes(N, M, edges);
# This code is contributed by rutvik_56
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
static int n = 10005;
// To store Prime Numbers
static bool []Prime = new bool[n + 1];
// Function to find the prime numbers
// till 10^5
static void SieveOfEratosthenes()
{
int i, j;
Prime[0] = Prime[1] = false;
for(i = 2; i * i <= 10005; i++)
{
// Traverse all multiple of i
// and make it false
if (Prime[i])
{
for(j = 2 * i; j < 10005; j += i)
{
Prime[j] = false;
}
}
}
}
// Function to print the nodes having
// prime degree
static void primeDegreeNodes(int N, int M,
int [,]edges)
{
// To store Adjacency List of
// a Graph
List []Adj = new List[N + 1];
for(int i = 0; i < Adj.Length; i++)
Adj[i] = new List();
// Make Adjacency List
for(int i = 0; i < M; i++)
{
int x = edges[i, 0];
int y = edges[i, 1];
Adj[x].Add(y);
Adj[y].Add(x);
}
// To precompute prime numbers
// till 10^5
SieveOfEratosthenes();
// Traverse each vertex
for(int i = 1; i <= N; i++)
{
// Find size of Adjacency List
int x = Adj[i].Count;
// If length of Adj[i] is Prime
// then print it
if (Prime[x])
Console.Write(i + " ");
}
}
// Driver code
public static void Main(String[] args)
{
// Vertices and Edges
int N = 4, M = 6;
// Edges
int [,]edges = { { 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 } };
for(int i = 0; i < Prime.Length; i++)
Prime[i] = true;
// Function Call
primeDegreeNodes(N, M, edges);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
1 2 3 4
时间复杂度: O(N + M) ,其中 N 是顶点数,M 是边数。
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