给定一个具有N个节点和N-1个边缘的无环无向图，其形式为2D数组arr [] [] ，其中每一行都由两个数字L和R组成，表示L和R之间的边缘。对于树中的每个节点X ，让dis(X)表示从X到最远节点的边数。任务是找到给定图的dis(x)的最小值。
例子：
Input: N = 6, arr[][] = { {1, 4}, {2, 3}, {3, 4}, {4, 5}, {5, 6} }
Output: 2
Explanation:
Below is the graph from the above information:
As we can see from the above graph the farthest node from vertex 0 is at distance 3. By repeating the DFS traversal for all the node in the graph, we have maximum distance[] from source node to farthest node as:
distance[] = {3, 4, 3, 2, 3, 4} and the minimum of the distances is the required result.
Input: N = 6, arr[][] = { {1, 2}, {1, 3}, {1, 4}, {2, 5}, {2, 6} }
Output: 2
Explanation:
The distance[] from every node to farthest node for the above graph is:
distance[] = {3, 4, 3, 2, 3, 4} and the minimum of the distances is 1.
方法：
想法是使用DFS遍历来解决此问题。步骤如下：
- 对于任何节点(例如a )，使用DFS遍历遍历图，节点自身的距离为0。
- 对于对节点a的每个递归调用，请使用数组中的节点a不断更新递归节点的距离(例如distance [] )。
- 通过在每次对Node a的递归调用中获取最大距离，可得出节点a与节点最远节点之间的边数。
- 对图中的所有节点重复上述步骤，并不断更新距离数组( distance [] )中每个节点的最远节点距离。
- 数组distance []的最小值是所需的结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Distance vector to find the distance of
// a node to it's farthest node
vector dist;
// To keep the track of visited array
// while DFS Traversal
vector vis;
// Function for DFS traversal to update
// the distance vector
void dfs(int u, vector Adj[], int s)
{
// Mark the visited array for vertex u
vis[u] = true;
// Traverse the adjacency list for u
for (auto& it : Adj[u]) {
// If the any node is not visited,
// then recursively call for next
// vertex with distance increment
// by 1
if (vis[it] == false) {
dfs(it, Adj, s + 1);
}
}
// Update the maximum distance for the
// farthest vertex from node u
dist[u] = max(dist[u], s);
}
// Function to find the minimum of the
// farthest vertex for every vertex in
// the graph
void minFarthestDistance(int arr[][2], int n)
{
// Resize distance vector
dist.resize(n + 1, 0);
// To create adjacency list for graph
vector Adj[n + 1];
// Create Adjacency list for every
// edge given in arr[][]
for (int i = 0; i < n - 1; i++) {
Adj[arr[i][0]].push_back(arr[i][1]);
Adj[arr[i][1]].push_back(arr[i][0]);
}
// DFS Traversal for every node in the
// graph to update the distance vector
for (int i = 1; i <= n; i++) {
// Clear and resize vis[] before
// DFS traversal for every vertex
vis.clear();
vis.resize(n + 1, false);
// DFS Traversal for vertex i
dfs(i, Adj, 0);
}
cout << *min_element(dist.begin() + 1,
dist.end());
}
// Driver Code
int main()
{
// Number of Nodes
int N = 6;
int arr[][2] = { { 1, 4 }, { 2, 3 }, { 3, 4 },
{ 4, 5 }, { 5, 6 } };
minFarthestDistance(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Distance vector to find the distance
// of a node to it's farthest node
static int[] dist;
// To keep the track of visited array
// while DFS Traversal
static boolean[] vis;
// Function for DFS traversal to update
// the distance vector
static void dfs(int u, Vector[] Adj, int s)
{
// Mark the visited array for vertex u
vis[u] = true;
// Traverse the adjacency list for u
for (int it : Adj[u])
{
// If the any node is not visited,
// then recursively call for next
// vertex with distance increment
// by 1
if (vis[it] == false)
{
dfs(it, Adj, s + 1);
}
}
// Update the maximum distance for
// the farthest vertex from node u
dist[u] = Math.max(dist[u], s);
}
// Function to find the minimum of the
// farthest vertex for every vertex in
// the graph
static void minFarthestDistance(int[][] arr, int n)
{
// Resize distance vector
dist = new int[n + 1];
Arrays.fill(dist, 0);
// To create adjacency list for graph
@SuppressWarnings("unchecked")
Vector[] Adj = new Vector[n + 1];
for(int i = 0; i < n + 1; i++)
{
Adj[i] = new Vector<>();
}
// Create Adjacency list for every
// edge given in arr[][]
for(int i = 0; i < n - 1; i++)
{
Adj[arr[i][0]].add(arr[i][1]);
Adj[arr[i][1]].add(arr[i][0]);
}
// DFS Traversal for every node in the
// graph to update the distance vector
for(int i = 1; i <= n; i++)
{
// Clear and resize vis[] before
// DFS traversal for every vertex
vis = new boolean[n + 1];
Arrays.fill(vis, false);
// DFS Traversal for vertex i
dfs(i, Adj, 0);
}
int min = Integer.MAX_VALUE;
for(int i = 1; i < dist.length; i++)
{
if (dist[i] < min)
min = dist[i];
}
System.out.println(min);
}
// Driver Code
public static void main(String[] args)
{
// Number of Nodes
int N = 6;
int[][] arr = { { 1, 4 }, { 2, 3 },
{ 3, 4 }, { 4, 5 },
{ 5, 6 } };
minFarthestDistance(arr, N);
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 program for the above approach
# Function for DFS traversal to update
# the distance vector
def dfs(u, s):
global vis, Adj, dist
# Mark the visited array for vertex u
vis[u] = True
# Traverse the adjacency list for u
for it in Adj[u]:
# If the any node is not visited,
# then recursively call for next
# vertex with distance increment
# by 1
if (vis[it] == False):
dfs(it, s + 1)
# Update the maximum distance for the
# farthest vertex from node u
dist[u] = max(dist[u], s)
# Function to find the minimum of the
# farthest vertex for every vertex in
# the graph
def minFarthestDistance(arr, n):
global dist, vis, Adj
# Create Adjacency list for every
# edge given in arr[][]
for i in range(n - 1):
Adj[arr[i][0]].append(arr[i][1])
Adj[arr[i][1]].append(arr[i][0])
# DFS Traversal for every node in the
# graph to update the distance vector
for i in range(1, n + 1):
# Clear and resize vis[] before
# DFS traversal for every vertex
# vis.clear()
for j in range(n + 1):
vis[j] = False
# vis.resize(n + 1, false)
# DFS Traversal for vertex i
dfs(i, 0)
print(min(dist[i] for i in range(1, n + 1)))
# Driver Code
if __name__ == '__main__':
dist = [0 for i in range(1001)]
vis = [False for i in range(1001)]
Adj = [[] for i in range(1001)]
# Number of Nodes
N = 6
arr = [ [ 1, 4 ], [ 2, 3 ],
[ 3, 4 ], [ 4, 5 ], [ 5, 6 ] ]
minFarthestDistance(arr, N)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Distance vector to find the distance
// of a node to it's farthest node
static int[] dist;
// To keep the track of visited array
// while DFS Traversal
static bool[] vis;
// Function for DFS traversal to update
// the distance vector
static void dfs(int u, List> Adj, int s)
{
// Mark the visited array for vertex u
vis[u] = true;
// Traverse the adjacency list for u
foreach(int it in Adj[u])
{
// If the any node is not visited,
// then recursively call for next
// vertex with distance increment
// by 1
if (vis[it] == false)
{
dfs(it, Adj, s + 1);
}
}
// Update the maximum distance for
// the farthest vertex from node u
dist[u] = Math.Max(dist[u], s);
}
// Function to find the minimum of the
// farthest vertex for every vertex in
// the graph
static void minFarthestDistance(int[,] arr, int n)
{
// Resize distance vector
dist = new int[n + 1];
Array.Fill(dist, 0);
// To create adjacency list for graph
List> Adj = new List>();
for(int i = 0; i < n + 1; i++)
{
Adj.Add(new List());
}
// Create Adjacency list for every
// edge given in arr[][]
for(int i = 0; i < n - 1; i++)
{
Adj[arr[i, 0]].Add(arr[i, 1]);
Adj[arr[i, 1]].Add(arr[i, 0]);
}
// DFS Traversal for every node in the
// graph to update the distance vector
for(int i = 1; i <= n; i++)
{
// Clear and resize vis[] before
// DFS traversal for every vertex
vis = new bool[n + 1];
Array.Fill(vis, false);
// DFS Traversal for vertex i
dfs(i, Adj, 0);
}
int min = Int32.MaxValue;
for(int i = 1; i < dist.Length; i++)
{
if (dist[i] < min)
{
min = dist[i];
}
}
Console.WriteLine(min);
}
// Driver Code
static public void Main ()
{
// Number of Nodes
int N = 6;
int[,] arr = { { 1, 4 }, { 2, 3 },{ 3, 4 },
{ 4, 5 }, { 5, 6 } };
minFarthestDistance(arr, N);
}
}
// This code is contributed by rag2127 输出：
2时间复杂度： O(V *(V + E))，其中V是顶点数，E是边数。