给定一个具有N 个节点和N-1 条边的连通无环图，找出彼此距离相等的节点对。
例子：
Input:
3
1 2
2 3
Output: 1
Explanation:
1
/
2
/
3
Input:
5
1 2
2 3
1 4
4 5
Output: 4方法：
- 假设一个图有 6 个级别（0 到 5），级别 0、2、4 的距离相等，但级别 1、3、5 的距离也是均匀的，因为它们的差异是 2，这是偶数，因此我们必须兼顾两者条件即计算偶数和奇数。
- 给定的问题可以通过执行 dfs 遍历来解决
- 选择任意一个源节点作为root，进行dfs遍历并维护访问过的节点
用于执行 dfs 和 dist 数组以计算与根的距离的数组 - 现在遍历距离数组并保持偶数级和奇数级的计数
- 计算总数为 ((even_count * (even_count-1)) + (odd_count * (odd_count-1))/2
下面是上述方法的实现：
C++
// C++ program to find
// the count of nodes
// at even distance
#include
using namespace std;
// Dfs function to find count of nodes at
// even distance
void dfs(vector graph[], int node, int dist[],
bool vis[], int c)
{
if (vis[node]) {
return;
}
// Set flag as true for current
// node in visited array
vis[node] = true;
// Insert the distance in
// dist array for current
// visited node u
dist[node] = c;
for (int i = 0; i < graph[node].size(); i++) {
// If its neighbours are not vis,
// run dfs for them
if (!vis[graph[node][i]]) {
dfs(graph, graph[node][i], dist, vis, c + 1);
}
}
}
int countOfNodes(vector graph[], int n)
{
// bool array to
// mark visited nodes
bool vis[n + 1] = { false };
// Integer array to
// compute distance
int dist[n + 1] = { 0 };
dfs(graph, 1, dist, vis, 0);
int even = 0, odd = 0;
// Traverse the distance array
// and count the even and odd levels
for (int i = 1; i <= n; i++) {
if (dist[i] % 2 == 0) {
even++;
}
else {
odd++;
}
}
int ans = ((even * (even - 1)) + (odd * (odd - 1))) / 2;
return ans;
}
// Driver code
int main()
{
int n = 5;
vector graph[n + 1] = { {},
{ 2 },
{ 1, 3 },
{ 2 } };
int ans = countOfNodes(graph, n);
cout << ans << endl;
return 0;
} Java
// Java program to find the count of
// nodes at even distance
import java.util.*;
class GFG
{
// Dfs function to find count of nodes at
// even distance
static void dfs(Vector graph[],
int node, int dist[],
boolean vis[], int c)
{
if (vis[node])
{
return;
}
// Set flag as true for current
// node in visited array
vis[node] = true;
// Insert the distance in
// dist array for current
// visited node u
dist[node] = c;
for (int i = 0; i < graph[node].size(); i++)
{
// If its neighbours are not vis,
// run dfs for them
if (!vis[graph[node].get(i)])
{
dfs(graph, graph[node].get(i),
dist, vis, c + 1);
}
}
}
static int countOfNodes(Vector graph[],
int n)
{
// bool array to
// mark visited nodes
boolean []vis = new boolean[n + 1];
// Integer array to
// compute distance
int []dist = new int[n + 1];
dfs(graph, 1, dist, vis, 0);
int even = 0, odd = 0;
// Traverse the distance array
// and count the even and odd levels
for (int i = 1; i <= n; i++)
{
if (dist[i] % 2 == 0)
{
even++;
}
else
{
odd++;
}
}
int ans = ((even * (even - 1)) +
(odd * (odd - 1))) / 2;
return ans;
}
// Driver code
public static void main(String[] args)
{
int n = 5;
Vector []graph = new Vector[n + 1];
for(int i = 0; i< n + 1; i++)
{
graph[i] = new Vector();
}
graph[0] = new Vector();
graph[1] = new Vector(Arrays.asList(2));
graph[2] = new Vector(1, 3);
graph[3] = new Vector(2);
int ans = countOfNodes(graph, n);
System.out.println(ans);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to find
# the count of nodes
# at even distance
# Dfs function to find count of
# nodes at even distance
def dfs(graph, node, dist, vis, c) :
if (vis[node]) :
return;
# Set flag as true for current
# node in visited array
vis[node] = True;
# Insert the distance in
# dist array for current
# visited node u
dist[node] = c;
for i in range(len(graph[node])) :
# If its neighbours are not vis,
# run dfs for them
if (not vis[graph[node][i]]) :
dfs(graph, graph[node][i],
dist, vis, c + 1);
def countOfNodes(graph, n) :
# bool array to
# mark visited nodes
vis = [False] * (n + 1);
# Integer array to
# compute distance
dist = [0] * (n + 1);
dfs(graph, 1, dist, vis, 0);
even = 0; odd = 0;
# Traverse the distance array
# and count the even and odd levels
for i in range(1, n + 1) :
if (dist[i] % 2 == 0) :
even += 1;
else :
odd += 1;
ans = ((even * (even - 1)) +
(odd * (odd - 1))) // 2;
return ans;
# Driver code
if __name__ == "__main__" :
n = 5;
graph = [[], [ 2 ], [ 1, 3 ], [ 2 ]];
ans = countOfNodes(graph, n);
print(ans);
# This code is contributed by kanugargngC#
// C# program to find the count of
// nodes at even distance
using System;
using System.Collections.Generic;
class GFG
{
// Dfs function to find count of
// nodes at even distance
static void dfs(List []graph,
int node, int []dist,
bool []vis, int c)
{
if (vis[node])
{
return;
}
// Set flag as true for current
// node in visited array
vis[node] = true;
// Insert the distance in
// dist array for current
// visited node u
dist[node] = c;
for (int i = 0; i < graph[node].Count; i++)
{
// If its neighbours are not vis,
// run dfs for them
if (!vis[graph[node][i]])
{
dfs(graph, graph[node][i],
dist, vis, c + 1);
}
}
}
static int countOfNodes(List []graph,
int n)
{
// bool array to
// mark visited nodes
bool []vis = new bool[n + 1];
// int array to
// compute distance
int []dist = new int[n + 1];
dfs(graph, 1, dist, vis, 0);
int even = 0, odd = 0;
// Traverse the distance array
// and count the even and odd levels
for (int i = 1; i <= n; i++)
{
if (dist[i] % 2 == 0)
{
even++;
}
else
{
odd++;
}
}
int ans = ((even * (even - 1)) +
(odd * (odd - 1))) / 2;
return ans;
}
// Driver code
public static void Main(String[] args)
{
int n = 5;
List []graph = new List[n + 1];
for(int i = 0; i< n + 1; i++)
{
graph[i] = new List();
}
graph[0] = new List{};
graph[1] = new List{2};
graph[2] = new List{1, 3};
graph[3] = new List{2};
int ans = countOfNodes(graph, n);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji Javascript
输出：
6时间复杂度： O(V+E)
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