查找无向图
中两个顶点之间是否存在路径
原文： https://www.geeksforgeeks.org/find-if-there-is-a-path-between-two-vertices-in-an-undirected-graph/
给定一个无向图，它具有N个顶点和E边以及该图中的两个顶点（U，V）， 检测两个顶点之间是否存在路径。 如果存在路径，则打印为“是” ，否则为为“否” 。
示例：
U = 1，V = 2 输出：
No说明： 两点之间没有边，因此无法从 2 到达 2 1。输入：
U = 1，V = 3 输出：
Yes说明：顶点 3 从顶点 1 经由顶点 2 或 4。
朴素的方法：
的想法是使用 Floyd Warshall 算法。 为了解决该问题，我们需要尝试所有范围为 [1，N] 的中间顶点，然后检查：
-
如果两个节点之间已经存在直接边。
-
或者，我们具有从节点
i到中间节点k的路径，以及从节点k到节点j的路径。
下面是上述方法的实现：
C++
// C++ program to detect if a path
// exists between any two vertices
// for the given undirected graph
#include <bits/stdc++.h>
using namespace std;
// Class representing a undirected
// graph using matrix
// representation
class Graph {
int V;
int** g;
public:
Graph(int V);
// Function to add an edge to graph
void addEdge(int v, int w);
// Function to check if
// there exists a path or not
bool isReachable(int s, int d);
// function to compute paths
// in the matrix using
// Floyd Warshall Algorithm
void computePaths();
};
Graph::Graph(int V)
{
this->V = V;
g = new int*[V + 1];
for (int i = 1; i < V + 1; i++) {
// Rows may not be contiguous
g[i] = new int[V + 1];
// Initialize all entries
// as false to indicate
// that there are
// no edges initially
memset(g[i], 0, (V + 1) * sizeof(int));
}
// Initializing node to itself
// as it is always reachable
for (int i = 1; i <= V; i++)
g[i][i] = 1;
}
// Function to add edge between nodes
void Graph::addEdge(int v, int w)
{
g[v][w] = 1;
g[w][v] = 1;
}
// Function to compute the path
void Graph::computePaths()
{
// Use Floyd Warshall algorithm
// to detect if a path exists
for (int k = 1; k <= V; k++) {
// Try every vertex as an
// intermediate vertex
// to check if a path exists
for (int i = 1; i <= V; i++) {
for (int j = 1; j <= V; j++)
g[i][j] = g[i][j]
| (g[i][k]
&& g[k][j]);
}
}
}
// Function to check if nodes are reachable
bool Graph::isReachable(int s, int d)
{
if (g[s][d] == 1)
return true;
else
return false;
}
// Driver code
int main()
{
Graph _g(4);
_g.addEdge(1, 2);
_g.addEdge(2, 3);
_g.addEdge(1, 4);
_g.computePaths();
int u = 4, v = 3;
if (_g.isReachable(u, v))
cout << "Yes\n";
else
cout << "No\n";
return 0;
}
Python
# Python3 program to detect if a path
# exists between any two vertices
# for the given undirected graph
# Class representing a undirected
# graph using matrix
# representation
class Graph:
def __init__(self, V):
self.V = V
# Initialize all entries
# as false to indicate
# that there are
# no edges initially
self.g = [[0 for j in range(self.V + 1)]
for i in range(self.V + 1)]
# Initializing node to itself
# as it is always reachable
for i in range(self.V + 1):
self.g[i][i] = 1
# Function to add edge between nodes
def addEdge(self, v, w):
self.g[v][w] = 1
self.g[w][v] = 1
# Function to compute the path
def computePaths(self):
# Use Floyd Warshall algorithm
# to detect if a path exists
for k in range(1, self.V + 1):
# Try every vertex as an
# intermediate vertex
# to check if a path exists
for i in range(1, self.V + 1):
for j in range(1, self.V + 1):
self.g[i][j] = (self.g[i][j] |
(self.g[i][k] and
self.g[k][j]))
# Function to check if nodes
# are reachable
def isReachable(self, s, d):
if (self.g[s][d] == 1):
return True
else:
return False
# Driver code
if __name__=='__main__':
_g = Graph(4)
_g.addEdge(1, 2)
_g.addEdge(2, 3)
_g.addEdge(1, 4)
_g.computePaths()
u = 4
v = 3
if (_g.isReachable(u, v)):
print('Yes')
else:
print('No')
# This code is contributed by rutvik_56
Output:
Yes
时间复杂度：O（V 3 ）
辅助空间：O（V 2 ）
高效解决方案，
我们可以使用 BFS 或 DFS 查找从 u 到 v 的路径。以下是基于 BFS 的解决方案
C++
// C++ program to check if there is exist a path between
// two vertices of an undirected graph.
#include <iostream>
#include <list>
using namespace std;
// This class represents an undirected graph
// using adjacency list representation
class Graph {
int V; // No. of vertices
// Pointer to an array containing adjacency lists
list<int>* adj;
public:
Graph(int V); // Constructor
// function to add an edge to graph
void addEdge(int v, int w);
bool isReachable(int s, int d);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
// A BFS based function to check whether d is reachable from s.
bool Graph::isReachable(int s, int d)
{
// Base case
if (s == d)
return true;
// Mark all the vertices as not visited
bool* visited = new bool[V];
for (int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list<int> queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// it will be used to get all adjacent vertices of a vertex
list<int>::iterator i;
while (!queue.empty()) {
// Dequeue a vertex from queue and print it
s = queue.front();
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it
// visited and enqueue it
for (i = adj[s].begin(); i != adj[s].end(); ++i) {
// If this adjacent node is the destination node,
// then return true
if (*i == d)
return true;
// Else, continue to do BFS
if (!visited[*i]) {
visited[*i] = true;
queue.push_back(*i);
}
}
}
// If BFS is complete without visiting d
return false;
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
int u = 1, v = 3;
if (g.isReachable(u, v))
cout << "\n There is a path from " << u << " to " << v;
else
cout << "\n There is no path from " << u << " to " << v;
return 0;
}
Output:
There is a path from 1 to 3
时间复杂度：O(V + E)
辅助空间：O(V)
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