C++ 程序用于找到无向图的联通分量

使用 DFS 可以找出无向图的弱联通或强联通分量。这是解决此问题的 C++ 程序。

使用的函数

Begin
Function fillorder() = fill stack with all the vertices.
   a) Mark the current node as visited and print it
   b) Recur for all the vertices adjacent to this vertex
   c) All vertices reachable from v are processed by now, push v to Stack
End
Begin
Function DFS() :
   a) Mark the current node as visited and print it
   b) Recur for all the vertices adjacent to this vertex
End

示例

#include <iostream>
#include <list>
#include <stack>
using namespace std;
class G {
   int m;
   list<int> *adj;
   //declaration of functions
   void fillOrder(int n, bool visited[], stack<int> &Stack);
   void DFS(int n, bool visited[]);
   public:
      G(int N); //constructor
      void addEd(int v, int w);
      int print();
      G getTranspose();
};
G::G(int m) {
   this->m = m;
   adj = new list<int> [m];
}
void G::DFS(int n, bool visited[]) {
   visited[n] = true; // Mark the current node as visited and print it
   cout << n << " ";
   list<int>::iterator i;
   //Recur for all the vertices adjacent to this vertex
   for (i = adj[n].begin(); i != adj[n].end(); ++i)
      if (!visited[*i])
         DFS(*i, visited);
}
G G::getTranspose() {
   G g(m);
   for (int n = 0; n< m; n++) {
      list<int>::iterator i;
      for (i = adj[n].begin(); i != adj[n].end(); ++i) {
         g.adj[*i].push_back(n);
      }
   }
   return g;
}
void G::addEd(int v, int w) {
   adj[v].push_back(w); //add w to v's list
}
void G::fillOrder(int v, bool visited[], stack<int> &Stack) {
   visited[v] = true; //Mark the current node as visited and print it
   list<int>::iterator i;
   //Recur for all the vertices adjacent to this vertex
   for (i = adj[v].begin(); i != adj[v].end(); ++i)
      if (!visited[*i])
         fillOrder(*i, visited, Stack);
         Stack.push(v);
}
int G::print() { //print the solution
   stack<int> Stack;
   bool *visited = new bool[m];
   for (int i = 0; i < m; i++)
      visited[i] = false;
      for (int i = 0; i < m; i++)
         if (visited[i] == false)
            fillOrder(i, visited, Stack);
   G graph= getTranspose(); //Create a reversed graph
   for (int i = 0; i < m; i++) //Mark all the vertices as not visited
      visited[i] = false;
   int count = 0;
   //now process all vertices in order defined by Stack
   while (Stack.empty() == false) {
      int v = Stack.top();
      Stack.pop(); //pop vertex from stack
      if (visited[v] == false) {
         graph.DFS(v, visited);
         cout << endl;
      }
      count++;
   }
   return count;
}
int main() {
   G g(5);
   g.addEd(2, 1);
   g.addEd(3, 2);
   g.addEd(1, 0);
   g.addEd(0, 3);
   g.addEd(3, 1);
   cout << "Following are strongly connected components
   in given graph \n";
   if (g.print() > 1) {
      cout << "Graph is weakly connected.";
   } else {
      cout << "Graph is strongly connected.";
   }
   return 0;
}

输出

Following are strongly connected components in given
graph
4
0 1 2 3
Graph is weakly connected.

Smita Kapse

更新于: 30-07-2019

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