母顶点：图 G = (V, E) 中的母顶点是一个顶点 v，使得 G 中的所有其他顶点都可以通过来自 v 的路径到达。 图中可以有零个、一个或多个母顶点一个图表。我们需要找到给定图中的所有母顶点。
例子 ：
Input : Given graph below
Output : 0 1 4
Explaination : In the given graph, the mother vertices are 0, 1 and 4 as there exists a path to each vertex from these vertices.
推荐：在继续解决方案之前，请先在 {IDE} 上尝试您的方法。
天真的方法：
一种简单的方法是在所有顶点上执行 DFS 或 BFS，并确定我们是否可以从该顶点到达所有顶点。
时间复杂度： O(V(E+V))
有效的方法：
- 使用此算法在给定的图 G 中找到任何母顶点v。
- 如果存在母顶点，则图 G 中形成强连通分量并包含v的顶点集是该图所有母顶点的集合。
上述想法如何运作？
如果图存在母顶点，则所有母顶点都是包含母顶点的强连通分量的顶点，因为如果v是母顶点并且存在从u -> v的路径，则u必须是母顶点顶点也是如此。
下面是上述方法的实现：
C++
// C++ program to find all the mother vertices
#include
using namespace std;
// This function does DFS traversal
// from given node u, and marks the
// visited nodes in the visited array
void dfs_helper(int u, vector >& adj,
bool visited[])
{
if (visited[u])
return;
visited[u] = true;
for (auto v : adj[u]) {
if (!visited[v])
dfs_helper(v, adj, visited);
}
}
// Function that stores the adjacency
// list of the transpose graph of the
// given graph in the trans_adj vector
void getTransposeGraph(vector >& adj,
vector >& trans_adj)
{
for (int u = 0; u < adj.size(); u++) {
for (auto v : adj[u]) {
trans_adj[v].push_back(u);
}
}
}
// Initializes all elements of visited
// array with value false
void initialize_visited(bool visited[], int n)
{
for (int u = 0; u < n; u++)
visited[u] = false;
}
// Returns the list of mother
// vertices. If the mother vertex
// does not exists returns -1
vector findAllMotherVertices(vector >& adj)
{
int n = adj.size();
bool visited[n];
// Find any mother vertex
// in given graph, G
initialize_visited(visited, n);
int last_dfs_called_on = -1;
for (int u = 0; u < n; u++) {
if (!visited[u]) {
dfs_helper(u, adj, visited);
last_dfs_called_on = u;
}
}
// Check if we can reach
// all vertices from
// last_dfs_called_on node
initialize_visited(visited, n);
dfs_helper(last_dfs_called_on, adj, visited);
for (int u = 0; u < n; u++) {
// Check of the mother vertex
// exist else return -1
if (!visited[u]) {
vector emptyVector;
emptyVector.push_back(-1);
return emptyVector;
}
}
// Now in G_transpose, do DFS
// from that mother vertex,
// and we will only reach the
// other mother vertices of G
int motherVertex = last_dfs_called_on;
// trans_adj is the transpose
// of the given Graph
vector > trans_adj(n);
// Function call to get
// the transpose graph
getTransposeGraph(adj, trans_adj);
// DFS from that mother vertex
// in the transpose graph and the
// visited nodes are all the
// mother vertices of the given
// graph G
initialize_visited(visited, n);
dfs_helper(motherVertex, trans_adj, visited);
// Vector to store the list
// of mother vertices
vector ans;
for (int u = 0; u < n; u++) {
if (visited[u])
ans.push_back(u);
}
// Return the required list
return ans;
}
// Driver Code
int main()
{
// No. of nodes
int V = 8;
vector > adj(V);
adj[0].push_back(1);
adj[1].push_back(2);
adj[1].push_back(4);
adj[1].push_back(5);
adj[2].push_back(3);
adj[2].push_back(6);
adj[3].push_back(2);
adj[3].push_back(7);
adj[4].push_back(0);
adj[4].push_back(5);
adj[5].push_back(6);
adj[6].push_back(5);
adj[6].push_back(7);
// Function call to find the mother vertces
vector motherVertices = findAllMotherVertices(adj);
// Print answer
if (motherVertices[0] == -1)
cout << "No mother vertex exists";
else {
cout << "All Mother vertices of the graph are : ";
for (int v : motherVertices)
cout << v << " ";
}
return 0;
} 输出
All Mother vertices of the graph are : 0 1 4 时间复杂度： O(V+E)
空间复杂度： O(V+E)
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。