给定一个无向图和两个顶点 X 和 Y ，我们的任务是检查顶点 X 是否位于顶点 Y 的子图中。
例子：
Input: X = 2, Y = 3
Output: No
Explanation:
Subgraph of a vertex Y = 3 is set of all the vertex which lies below Y and are reachable by Y. Here subgraph of 3 contains {6} not 2.
Input: X = 6, Y = 1
Output: Yes
Explanation:
Subgraph of 1 contain {2, 3, 4, 5, 6} so 6 lies in subgraph of 1.
方法：想法是使用深度优先搜索(DFS)。初始化两个数组in和out用于维护遍历顶点的开始时间和结束时间以标记直到遍历顶点。如果第二个顶点的开始时间小于第一个顶点的开始时间并且第一个顶点的结束时间小于第二个顶点的结束时间则返回真否则返回假。
下面是上述方法的实现：
C++
// C++ implementation to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
#include
using namespace std;
int cnt = 1;
// Function ot perform dfs
void dfs(vector v[], int in[],
int out[], int visited[], int i)
{
// Mark visited of vertex i
visited[i] = 1;
// Update starting time
// of vertex i
in[i] = cnt;
// Increment the cnt
cnt++;
for (auto x : v[i]) {
// Check if not visited
// call dfs from x
if (!visited[x])
dfs(v, in, out, visited, x);
}
// Update ending time
// of vertex i
out[i] = cnt;
// Increment the cnt
cnt++;
}
// Function to add edges in graph
void addedge(vector v[], int x, int y)
{
v[x].push_back(y);
v[y].push_back(x);
}
// Function to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
bool is_subtree(vector v[], int n,
int m, int x, int y)
{
// Arrays for starting time,
// ending time and to check
// for visited respectively
int in[n + 1], out[n + 1], visited[n + 1];
// Mark all vertices starting time,
// ending time and visited as zero
for (int i = 1; i <= n; i++) {
in[i] = 0;
out[i] = 0;
visited[i] = 0;
}
// Check if y comes before x
// and leaves after x then x lies
// in the subgraph of y
// call dfs from any vertex,
// here we have called from 1
dfs(v, in, out, visited, 1);
if (in[y] < in[x] && out[y] > out[x])
return true;
else
return false;
}
// Driver code
int main()
{
// n number of vertices
// m number of edges
int n = 6, m = 5;
// Create a graph given
// in the above diagram
vector v[n + 1];
addedge(v, 1, 2);
addedge(v, 1, 3);
addedge(v, 2, 4);
addedge(v, 1, 5);
addedge(v, 3, 6);
int x = 6, y = 1;
if (is_subtree(v, n, m, x, y))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
import java.util.*;
class GFG{
static int cnt = 1;
// Function ot perform dfs
static void dfs(Vector v[], int in[],
int out[], int visited[], int i)
{
// Mark visited of vertex i
visited[i] = 1;
// Update starting time
// of vertex i
in[i] = cnt;
// Increment the cnt
cnt++;
for(int x : v[i])
{
// Check if not visited
// call dfs from x
if (visited[x] == 0)
dfs(v, in, out, visited, x);
}
// Update ending time
// of vertex i
out[i] = cnt;
// Increment the cnt
cnt++;
}
// Function to add edges in graph
static void addedge(Vector v[],
int x, int y)
{
v[x].add(y);
v[y].add(x);
}
// Function to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
static boolean is_subtree(Vector v[],
int n, int m, int x,
int y)
{
// Arrays for starting time,
// ending time and to check
// for visited respectively
int []in = new int[n + 1];
int []out = new int[n + 1];
int []visited = new int[n + 1];
// Mark all vertices starting time,
// ending time and visited as zero
for(int i = 1; i <= n; i++)
{
in[i] = 0;
out[i] = 0;
visited[i] = 0;
}
// Check if y comes before x
// and leaves after x then x lies
// in the subgraph of y
// call dfs from any vertex,
// here we have called from 1
dfs(v, in, out, visited, 1);
if (in[y] < in[x] && out[y] > out[x])
return true;
else
return false;
}
// Driver code
public static void main(String[] args)
{
// n number of vertices
// m number of edges
int n = 6, m = 5;
// Create a graph given
// in the above diagram
@SuppressWarnings("unchecked")
Vector []v = new Vector[n + 1];
for(int i = 0; i < v.length; i++)
v[i] = new Vector();
addedge(v, 1, 2);
addedge(v, 1, 3);
addedge(v, 2, 4);
addedge(v, 1, 5);
addedge(v, 3, 6);
int x = 6, y = 1;
if (is_subtree(v, n, m, x, y))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation to check if
# vertex X lies in subgraph of
# vertex Y for the given graph
cnt = 1
# Function to perform dfs
def dfs(v, in_, out, visited, i):
global cnt
# Mark visited of vertex i
visited[i] = 1
# Update starting time
# of vertex i
in_[i] = cnt
# Increment the cnt
cnt += 1
# Check if not visited
# call dfs from x
for x in v[i]:
if not visited[x]:
dfs(v, in_, out, visited, x)
# Update ending time
# of vertex i
out[i] = cnt
# Increment the cnt
cnt += 1
# Function to add edges in graph
def addedge(v, x, y):
v[x].append(y)
v[y].append(x)
# Function to check if vertex X
# lies in subgraph of vertex Y
# for the given graph
def is_subtree(v, n, m, x, y):
# Arrays for starting time,
# ending time and to check
# for visited respectively
# Mark all vertices starting time,
# ending time and visited as zero
in_ = [0] * (n + 1)
out = [0] * (n + 1)
visited = [0] * (n + 1)
# Check if y comes before x
# and leaves after x then x lies
# in the subgraph of y
# call dfs from any vertex,
# here we have called from 1
dfs(v, in_, out, visited, 1)
if in_[y] < in_[x] and out[y] > out[x]:
return True
else:
return False
# Driver code
# n number of vertices
# m number of edges
n, m = 6, 5
# Create a graph given
# in the above diagram
v = []
for i in range(n + 1):
v.append([])
addedge(v, 1, 2)
addedge(v, 1, 3)
addedge(v, 2, 4)
addedge(v, 1, 5)
addedge(v, 3, 6)
x, y = 6, 1
if is_subtree(v, n, m, x, y):
print("Yes")
else:
print("No")
# This code is contributed by Stuti PathakC#
// C# implementation to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
using System;
using System.Collections.Generic;
class GFG{
static int cnt = 1;
// Function ot perform dfs
static void dfs(List []v, int []In,
int []Out, int []visited, int i)
{
// Mark visited of vertex i
visited[i] = 1;
// Update starting time
// of vertex i
In[i] = cnt;
// Increment the cnt
cnt++;
foreach(int x in v[i])
{
// Check if not visited
// call dfs from x
if (visited[x] == 0)
dfs(v, In, Out, visited, x);
}
// Update ending time
// of vertex i
Out[i] = cnt;
// Increment the cnt
cnt++;
}
// Function to add edges in graph
static void addedge(List []v,
int x, int y)
{
v[x].Add(y);
v[y].Add(x);
}
// Function to check if vertex X
// lies in subgraph of vertex Y
// for the given graph
static bool is_subtree(List []v,
int n, int m,
int x, int y)
{
// Arrays for starting time,
// ending time and to check
// for visited respectively
int []In = new int[n + 1];
int []Out = new int[n + 1];
int []visited = new int[n + 1];
// Mark all vertices starting time,
// ending time and visited as zero
for(int i = 1; i <= n; i++)
{
In[i] = 0;
Out[i] = 0;
visited[i] = 0;
}
// Check if y comes before x
// and leaves after x then x lies
// in the subgraph of y
// call dfs from any vertex,
// here we have called from 1
dfs(v, In, Out, visited, 1);
if (In[y] < In[x] && Out[y] > Out[x])
return true;
else
return false;
}
// Driver code
public static void Main(String[] args)
{
// n number of vertices
// m number of edges
int n = 6, m = 5;
// Create a graph given
// in the above diagram
List []v = new List[n + 1];
for(int i = 0; i < v.Length; i++)
v[i] = new List();
addedge(v, 1, 2);
addedge(v, 1, 3);
addedge(v, 2, 4);
addedge(v, 1, 5);
addedge(v, 3, 6);
int x = 6, y = 1;
if (is_subtree(v, n, m, x, y))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Rohit_ranjan 输出：
Yes
时间复杂度： O(V + E)
空间复杂度： O(3*N)
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