给定一个由n个顶点和m个边组成的无向无加权连通图。任务是找到此图的任何生成树，以使所有顶点上的最大度数最大可能。打印输出边的顺序无关紧要，也可以反向打印边，即(u，v)也可以打印为(v，u)。
例子：
Input:
1
/ \
2 5
\ /
3
|
4
Output:
3 2
3 5
3 4
1 2
The maximum degree over all vertices
is of vertex 3 which is 3 and is
maximum possible.
Input:
1
/
2
/ \
5 3
|
4
Output:
2 1
2 5
2 3
3 4先决条件： Kruskal算法来查找最小生成树
方法：可以使用Kruskal算法找到最小生成树来解决给定的问题。
我们在图中找到最大度数的顶点。首先，我们将执行与该顶点有关的所有边的合并，然后执行正常的Kruskal算法。这为我们提供了最佳的生成树。
Java
// Java implementation of the approach
import java.util.*;
public class GFG {
// par and rank will store the parent
// and rank of particular node
// in the Union Find Algorithm
static int par[], rank[];
// Find function of Union Find Algorithm
static int find(int x)
{
if (par[x] != x)
par[x] = find(par[x]);
return par[x];
}
// Union function of Union Find Algorithm
static void union(int u, int v)
{
int x = find(u);
int y = find(v);
if (x == y)
return;
if (rank[x] > rank[y])
par[y] = x;
else if (rank[x] < rank[y])
par[x] = y;
else {
par[x] = y;
rank[y]++;
}
}
// Function to find the required spanning tree
static void findSpanningTree(int deg[], int n,
int m, ArrayList g[])
{
par = new int[n + 1];
rank = new int[n + 1];
// Initialising parent of a node
// by itself
for (int i = 1; i <= n; i++)
par[i] = i;
// Variable to store the node
// with maximum degree
int max = 1;
// Finding the node with maximum degree
for (int i = 2; i <= n; i++)
if (deg[i] > deg[max])
max = i;
// Union of all edges incident
// on vertex with maximum degree
for (int v : g[max]) {
System.out.println(max + " " + v);
union(max, v);
}
// Carrying out normal Kruskal Algorithm
for (int u = 1; u <= n; u++) {
for (int v : g[u]) {
int x = find(u);
int y = find(v);
if (x == y)
continue;
union(x, y);
System.out.println(u + " " + v);
}
}
}
// Driver code
public static void main(String args[])
{
// Number of nodes
int n = 5;
// Number of edges
int m = 5;
// ArrayList to store the graph
ArrayList g[] = new ArrayList[n + 1];
for (int i = 1; i <= n; i++)
g[i] = new ArrayList<>();
// Array to store the degree
// of each node in the graph
int deg[] = new int[n + 1];
// Add edges and update degrees
g[1].add(2);
g[2].add(1);
deg[1]++;
deg[2]++;
g[1].add(5);
g[5].add(1);
deg[1]++;
deg[5]++;
g[2].add(3);
g[3].add(2);
deg[2]++;
deg[3]++;
g[5].add(3);
g[3].add(5);
deg[3]++;
deg[5]++;
g[3].add(4);
g[4].add(3);
deg[3]++;
deg[4]++;
findSpanningTree(deg, n, m, g);
}
} Python3
# Python3 implementation of the approach
from typing import List
# par and rank will store the parent
# and rank of particular node
# in the Union Find Algorithm
par = []
rnk = []
# Find function of Union Find Algorithm
def find(x: int) -> int:
global par
if (par[x] != x):
par[x] = find(par[x])
return par[x]
# Union function of Union Find Algorithm
def Union(u: int, v: int) -> None:
global par, rnk
x = find(u)
y = find(v)
if (x == y):
return
if (rnk[x] > rnk[y]):
par[y] = x
elif (rnk[x] < rnk[y]):
par[x] = y
else:
par[x] = y
rnk[y] += 1
# Function to find the required spanning tree
def findSpanningTree(deg: List[int], n: int, m: int,
g: List[List[int]]) -> None:
global rnk, par
# Initialising parent of a node
# by itself
par = [i for i in range(n + 1)]
rnk = [0] * (n + 1)
# Variable to store the node
# with maximum degree
max = 1
# Finding the node with maximum degree
for i in range(2, n + 1):
if (deg[i] > deg[max]):
max = i
# Union of all edges incident
# on vertex with maximum degree
for v in g[max]:
print("{} {}".format(max, v))
Union(max, v)
# Carrying out normal Kruskal Algorithm
for u in range(1, n + 1):
for v in g[u]:
x = find(u)
y = find(v)
if (x == y):
continue
Union(x, y)
print("{} {}".format(u, v))
# Driver code
if __name__ == "__main__":
# Number of nodes
n = 5
# Number of edges
m = 5
# ArrayList to store the graph
g = [[] for _ in range(n + 1)]
# Array to store the degree
# of each node in the graph
deg = [0] * (n + 1)
# Add edges and update degrees
g[1].append(2)
g[2].append(1)
deg[1] += 1
deg[2] += 1
g[1].append(5)
g[5].append(1)
deg[1] += 1
deg[5] += 1
g[2].append(3)
g[3].append(2)
deg[2] += 1
deg[3] += 1
g[5].append(3)
g[3].append(5)
deg[3] += 1
deg[5] += 1
g[3].append(4)
g[4].append(3)
deg[3] += 1
deg[4] += 1
findSpanningTree(deg, n, m, g)
# This code is contributed by sanjeev2552C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// par and rank will store the parent
// and rank of particular node
// in the Union Find Algorithm
static int []par;
static int []rank;
// Find function of Union Find Algorithm
static int find(int x)
{
if (par[x] != x)
par[x] = find(par[x]);
return par[x];
}
// Union function of Union Find Algorithm
static void union(int u, int v)
{
int x = find(u);
int y = find(v);
if (x == y)
return;
if (rank[x] > rank[y])
par[y] = x;
else if (rank[x] < rank[y])
par[x] = y;
else {
par[x] = y;
rank[y]++;
}
}
// Function to find the required spanning tree
static void findSpanningTree(int []deg, int n,
int m, List []g)
{
par = new int[n + 1];
rank = new int[n + 1];
// Initialising parent of a node
// by itself
for (int i = 1; i <= n; i++)
par[i] = i;
// Variable to store the node
// with maximum degree
int max = 1;
// Finding the node with maximum degree
for (int i = 2; i <= n; i++)
if (deg[i] > deg[max])
max = i;
// Union of all edges incident
// on vertex with maximum degree
foreach (int v in g[max])
{
Console.WriteLine(max + " " + v);
union(max, v);
}
// Carrying out normal Kruskal Algorithm
for (int u = 1; u <= n; u++)
{
foreach (int v in g[u])
{
int x = find(u);
int y = find(v);
if (x == y)
continue;
union(x, y);
Console.WriteLine(u + " " + v);
}
}
}
// Driver code
public static void Main(String []args)
{
// Number of nodes
int n = 5;
// Number of edges
int m = 5;
// ArrayList to store the graph
List []g = new List[n + 1];
for (int i = 1; i <= n; i++)
g[i] = new List();
// Array to store the degree
// of each node in the graph
int []deg = new int[n + 1];
// Add edges and update degrees
g[1].Add(2);
g[2].Add(1);
deg[1]++;
deg[2]++;
g[1].Add(5);
g[5].Add(1);
deg[1]++;
deg[5]++;
g[2].Add(3);
g[3].Add(2);
deg[2]++;
deg[3]++;
g[5].Add(3);
g[3].Add(5);
deg[3]++;
deg[5]++;
g[3].Add(4);
g[4].Add(3);
deg[3]++;
deg[4]++;
findSpanningTree(deg, n, m, g);
}
}
// This code has been contributed by 29AjayKumar 输出：
3 2
3 5
3 4
1 2如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。