给定一棵树，该树具有由N-1个边缘连接的N个节点和一个断开连接的节点，任务是在将给定树与给定的断开连接的零件连接后，找到给定树的每个节点的直径。
例子：
Input:
Output: 3 3 4 4 4 4
Explanation:
Initially diameter of tree = 3.
If an edge is added between node 1 and node 7, then the diameter of the tree is equal to 3(7 -> 1 -> 2 -> 5).
If an edge is added between node 2 and node 7, then the diameter of the tree is equal to 3(3 -> 1 -> 2 -> 5).
If an edge is added between node 3 and node 7, then the diameter of the tree is equal to 4(7 -> 3 -> 1 -> 2 -> 5).
If an edge is added between node 4 and node 7, then the diameter of the new tree is equal to 4(7-> 4 -> 2 -> 1 -> 3).
If an edge is added between node 5 and node 7, then the diameter of the new tree will be 4(7-> 5 -> 2 -> 1 -> 3).
If an edge is added between node 6 and node 7, then the diameter of the new tree will be 4((7-> 6 -> 2 -> 1 -> 3)).
Input:
Output: 3 2 3
Explanation:
Initial diameter of the tree = 2
If an edge is added between node 1 and node 4, then the diameter of the tree is equal to 3(4 -> 1 -> 2 -> 3).
If an edge is added between node 2 and node 4, then the diameter of the tree is equal to 2(4 -> 2 -> 3).
If an edge is added between node 3 and node 4, then the diameter of the tree is equal to 3(4 -> 3 -> 2 -> 1).
方法：要解决此问题，需要进行以下观察：
- The diameter increases by 1 when the disconnected node is connected to an edge which forms the end of the diameter.
- For all other nodes, the diameter remains unchanged on connecting the disconnected node.
请根据以下观察结果，按照以下步骤解决问题：
- 执行给定树的DFS遍历。
- 遍历时，请记录所经过的最远距离和最远的节点。
- 现在，从上述步骤获得的最远节点执行DFS ，并跟踪距该节点最远的节点。
- 现在，执行DFS，并继续在距上述步骤中获得的两个节点最远的Map中添加节点。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Keeps track of the farthest
// end of the diameter
int X = 1;
// Keeps track of the length of
// the diameter
int diameter = 0;
// Stores the nodes which are at
// ends of the diameter
map mp;
// Perform DFS on the given tree
void dfs(int current_node, int prev_node,
int len, bool add_to_map,
vector >& adj)
{
// Update diameter and X
if (len > diameter) {
diameter = len;
X = current_node;
}
// If current node is an end of diameter
if (add_to_map && len == diameter) {
mp[current_node] = 1;
}
// Traverse its neighbors
for (auto& it : adj[current_node]) {
if (it != prev_node)
dfs(it, current_node, len + 1,
add_to_map, adj);
}
}
// Function to call DFS for the
// required purposes
void dfsUtility(vector >& adj)
{
// DFS from a random node and find
// the node farthest from it
dfs(1, -1, 0, 0, adj);
int farthest_node = X;
// DFS from X to calculate diameter
dfs(farthest_node, -1, 0, 0, adj);
// DFS from farthest_node to find
// the farthest node(s) from it
dfs(farthest_node, -1, 0, 1, adj);
// DFS from X (other end of diameter) and
// check the farthest node(s) from it
dfs(X, -1, 0, 1, adj);
}
void printDiameters(vector >& adj)
{
dfsUtility(adj);
for (int i = 1; i <= 6; i++) {
// If node i is the end of
// a diameter
if (mp[i] == 1)
// Increase diameter by 1
cout << diameter + 1 << ", ";
// Otherwise
else
// Remains unchanged
cout << diameter << ", ";
}
}
// Driver Code
int main()
{
/* constructed tree is
1
/ \
2 3 7
/|\
/ | \
4 5 6 */
vector > adj(7);
// creating undirected edges
adj[1].push_back(2);
adj[2].push_back(1);
adj[1].push_back(3);
adj[3].push_back(1);
adj[2].push_back(4);
adj[4].push_back(2);
adj[2].push_back(5);
adj[5].push_back(2);
adj[2].push_back(6);
adj[6].push_back(2);
printDiameters(adj);
return 0;
} Java
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Keeps track of the farthest
// end of the diameter
static int X = 1;
// Keeps track of the length of
// the diameter
static int diameter = 0;
// Stores the nodes which are at
// ends of the diameter
static HashMap mp = new HashMap<>();
// Perform DFS on the given tree
static void dfs(int current_node, int prev_node,
int len, boolean add_to_map,
Vector [] adj)
{
// Update diameter and X
if (len > diameter)
{
diameter = len;
X = current_node;
}
// If current node is an end of diameter
if (add_to_map && len == diameter)
{
mp.put(current_node, true);
}
// Traverse its neighbors
for (int it : adj[current_node])
{
if (it != prev_node)
dfs(it, current_node, len + 1,
add_to_map, adj);
}
}
// Function to call DFS for the
// required purposes
static void dfsUtility(Vector [] adj)
{
// DFS from a random node and find
// the node farthest from it
dfs(1, -1, 0, false, adj);
int farthest_node = X;
// DFS from X to calculate diameter
dfs(farthest_node, -1, 0, false, adj);
// DFS from farthest_node to find
// the farthest node(s) from it
dfs(farthest_node, -1, 0, true, adj);
// DFS from X (other end of diameter) and
// check the farthest node(s) from it
dfs(X, -1, 0, true, adj);
}
static void printDiameters(Vector [] adj)
{
dfsUtility(adj);
for (int i = 1; i <= 6; i++)
{
// If node i is the end of
// a diameter
if (mp.containsKey(i) &&
mp.get(i) == true)
// Increase diameter by 1
System.out.print(diameter + 1 + ", ");
// Otherwise
else
// Remains unchanged
System.out.print(diameter + ", ");
}
}
// Driver Code
public static void main(String[] args)
{
/* constructed tree is
1
/ \
2 3 7
/|\
/ | \
4 5 6 */
Vector []adj = new Vector[7];
for (int i = 0; i < adj.length; i++)
adj[i] = new Vector();
// creating undirected edges
adj[1].add(2);
adj[2].add(1);
adj[1].add(3);
adj[3].add(1);
adj[2].add(4);
adj[4].add(2);
adj[2].add(5);
adj[5].add(2);
adj[2].add(6);
adj[6].add(2);
printDiameters(adj);
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program to implement
# the above approach
from collections import defaultdict
# Keeps track of the farthest
# end of the diameter
X = 1
# Keeps track of the length of
# the diameter
diameter = 0
# Stores the nodes which are at
# ends of the diameter
mp = defaultdict(lambda :0)
# Perform DFS on the given tree
def dfs(current_node, prev_node,
len, add_to_map, adj):
global diameter, X
# Update diameter and X
if len > diameter:
diameter = len
X = current_node
# If current node is an end of diameter
if add_to_map and len == diameter:
mp[current_node] = 1
# Traverse its neighbors
for it in adj[current_node]:
if it != prev_node:
dfs(it, current_node, len + 1,
add_to_map, adj)
# Function to call DFS for the
# required purposes
def dfsUtility(adj):
# DFS from a random node and find
# the node farthest from it
dfs(1, -1, 0, 0, adj)
farthest_node = X
# DFS from X to calculate diameter
dfs(farthest_node, -1, 0, 0, adj)
# DFS from farthest_node to find
# the farthest node(s) from it
dfs(farthest_node, -1, 0, 1, adj)
# DFS from X (other end of diameter) and
# check the farthest node(s) from it
dfs(X, -1, 0, 1, adj)
def printDiameters(adj):
global diameter
dfsUtility(adj)
for i in range(1, 6 + 1):
# If node i is the end of
# a diameter
if mp[i] == 1:
# Increase diameter by 1
print(diameter + 1, end = ", ")
# Otherwise
else:
# Remains unchanged
print(diameter, end = ", ")
# Driver code
# constructed tree is
# 1
# / \
# 2 3
# / | \
# 4 5 6
# |
# 7
adj = []
for i in range(7):
adj.append([])
# Creating undirected edges
adj[1].append(2)
adj[2].append(1)
adj[1].append(3)
adj[3].append(1)
adj[2].append(4)
adj[4].append(2)
adj[2].append(5)
adj[5].append(2)
adj[2].append(6)
adj[6].append(2)
printDiameters(adj)
# This code is contributed by Stuti PathakC#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
// Keeps track of the farthest
// end of the diameter
static int X = 1;
// Keeps track of the
// length of the diameter
static int diameter = 0;
// Stores the nodes which are at
// ends of the diameter
static Dictionary mp
= new Dictionary();
// Perform DFS on the given tree
static void dfs(int current_node, int prev_node,
int len, bool add_to_map,
List[] adj)
{
// Update diameter and X
if (len > diameter)
{
diameter = len;
X = current_node;
}
// If current node is an end of diameter
if (add_to_map && len == diameter)
{
mp.Add(current_node, true);
}
// Traverse its neighbors
foreach(int it in adj[current_node])
{
if (it != prev_node)
dfs(it, current_node, len + 1,
add_to_map, adj);
}
}
// Function to call DFS for
// the required purposes
static void dfsUtility(List[] adj)
{
// DFS from a random node and find
// the node farthest from it
dfs(1, -1, 0, false, adj);
int farthest_node = X;
// DFS from X to calculate diameter
dfs(farthest_node, -1, 0, false, adj);
// DFS from farthest_node to find
// the farthest node(s) from it
dfs(farthest_node, -1, 0, true, adj);
// DFS from X (other end of diameter) and
// check the farthest node(s) from it
dfs(X, -1, 0, true, adj);
}
static void printDiameters(List[] adj)
{
dfsUtility(adj);
for (int i = 1; i <= 6; i++)
{
// If node i is the end
// of a diameter
if (mp.ContainsKey(i) && mp[i] == true)
// Increase diameter by 1
Console.Write(diameter + 1 + ", ");
// Otherwise
else
// Remains unchanged
Console.Write(diameter + ", ");
}
}
// Driver Code
public static void Main(String[] args)
{
/* constructed tree is
1
/ \
2 3 7
/|\
/ | \
4 5 6 */
List[] adj = new List[ 7 ];
for (int i = 0; i < adj.Length; i++)
adj[i] = new List();
// creating undirected edges
adj[1].Add(2);
adj[2].Add(1);
adj[1].Add(3);
adj[3].Add(1);
adj[2].Add(4);
adj[4].Add(2);
adj[2].Add(5);
adj[5].Add(2);
adj[2].Add(6);
adj[6].Add(2);
printDiameters(adj);
}
}
// This code is contributed by Amit Katiyar 输出：
3, 3, 4, 4, 4, 4,
时间复杂度： O(V + E) ，其中V是顶点数，E是图形中边数。
辅助空间： O(V)