给定一个具有N个节点和M个边的图。任务是找到从给定图的每个连接的组件中选择节点的方法。
例子：
Input:
Output: 3
(1, 4), (2, 4), (3, 4) are possible ways.
Input:
Output: 6
(1, 4, 5), (2, 4, 5), (3, 4, 5), (1, 4, 6), (2, 4, 6), (3, 4, 6) are possible ways.
方法：每个连接组件中节点数的乘积是必需的答案。运行一个简单的dfs来查找每个连接的组件中的节点数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define N 100005
int n, m, temp;
vector gr[N];
int vis[N];
// Function to add edges in the graph
void Add_edges(int x, int y)
{
gr[x].push_back(y);
gr[y].push_back(x);
}
// Function for DFS
void dfs(int ch)
{
// Mark node as visited
vis[ch] = 1;
// Count number of nodes in a component
temp++;
for (auto i : gr[ch])
if (!vis[i])
dfs(i);
}
// Function to return the required number of ways
int NumberOfWays()
{
// To store the required answer
int ans = 1;
memset(vis, 0, sizeof vis);
for (int i = 1; i <= n; i++) {
// If current node hasn't been visited yet
if (!vis[i]) {
temp = 0;
dfs(i);
// Multiply it with the answer
ans *= temp;
}
}
return ans;
}
// Driver code
int main()
{
n = 4, m = 2;
// Add edges
Add_edges(1, 2);
Add_edges(1, 3);
cout << NumberOfWays();
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static final int N = 100005;
static int n, m, temp;
static Vector []gr = new Vector[N];
static int []vis = new int[N];
// Function to add edges in the graph
static void Add_edges(int x, int y)
{
gr[x].add(y);
gr[y].add(x);
}
// Function for DFS
static void dfs(int ch)
{
// Mark node as visited
vis[ch] = 1;
// Count number of nodes in a component
temp++;
for (int i : gr[ch])
if (vis[i] == 0)
dfs(i);
}
// Function to return the required number of ways
static int NumberOfWays()
{
// To store the required answer
int ans = 1;
Arrays.fill(vis, 0);
for (int i = 1; i <= n; i++)
{
// If current node hasn't been visited yet
if (vis[i] == 0)
{
temp = 0;
dfs(i);
// Multiply it with the answer
ans *= temp;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
n = 4;
m = 2;
for (int i = 0; i < N; i++)
gr[i] = new Vector();
// Add edges
Add_edges(1, 2);
Add_edges(1, 3);
System.out.print(NumberOfWays());
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation of the approach
# Function to add edges in the graph
def Add_edges(x, y):
gr[x].append(y)
gr[y].append(x)
# Function for DFS
def dfs(ch):
# Mark node as visited
vis[ch] = 1
global temp
# Count number of nodes
# in a component
temp += 1
for i in gr[ch]:
if not vis[i]:
dfs(i)
# Function to return the required
# number of ways
def NumberOfWays():
# To store the required answer
ans = 1
global temp
for i in range(1, n + 1):
# If current node hasn't been
# visited yet
if not vis[i]:
temp = 0
dfs(i)
# Multiply it with the answer
ans *= temp
return ans
# Driver code
if __name__ == "__main__":
n, m, temp = 4, 2, 0
N = 100005
gr = [[] for i in range(N)]
vis = [None] * N
# Add edges
Add_edges(1, 2)
Add_edges(1, 3)
print(NumberOfWays())
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static readonly int N = 100005;
static int n, m, temp;
static List []gr = new List[N];
static int []vis = new int[N];
// Function to add edges in the graph
static void Add_edges(int x, int y)
{
gr[x].Add(y);
gr[y].Add(x);
}
// Function for DFS
static void dfs(int ch)
{
// Mark node as visited
vis[ch] = 1;
// Count number of nodes in a component
temp++;
foreach (int i in gr[ch])
if (vis[i] == 0)
dfs(i);
}
// Function to return the required number of ways
static int NumberOfWays()
{
// To store the required answer
int ans = 1;
for(int i= 0; i < N; i++)
vis[i] = 0;
for (int i = 1; i <= n; i++)
{
// If current node hasn't been visited yet
if (vis[i] == 0)
{
temp = 0;
dfs(i);
// Multiply it with the answer
ans *= temp;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
n = 4;
m = 2;
for (int i = 0; i < N; i++)
gr[i] = new List();
// Add edges
Add_edges(1, 2);
Add_edges(1, 3);
Console.Write(NumberOfWays());
}
}
// This code is contributed by 29AjayKumar 输出：
3