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法律研究针对有向图检查其是否是弱连通或强连通的 C++ 程序
可使用 DFS 找出针对给定有向图的弱连通还是强连通。这是一个针对此问题的 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.
更新于： 30-Jul-2019
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