Question

找不到答案.... mea culpea。

Answer 1

这是一个查找无向图的连通分量的程序。这可以通过深度优先搜索使用它作为成功开发和完成你的基准来完成。

// Implementation of Kosaraju's algorithm to print all SCCs
#include <iostream>
#include <list>
#include <stack>
using namespace std;

class Graph{
    int V; // No. of vertices
    list<int> *adj; // An array of adjacency lists

    // Fills Stack with vertices (in increasing order of finishing times)
    // The top element of stack has the maximum finishing time
    void fillOrder(int v, bool visited[], stack<int> &Stack);

    // A recursive function to print DFS starting from v
    void DFSUtil(int v, bool visited[]);
public:
    Graph(int V);
    void addEdge(int v, int w);

    // The main function that finds and prints strongly connected components
    int printSCCs();

    // Function that returns reverse (or transpose) of this graph
    Graph getTranspose();
};

Graph::Graph(int V){
    this->V = V;
    adj = new list<int> [V];
}

// A recursive function to print DFS starting from v
void Graph::DFSUtil(int v, bool visited[]){
    // Mark the current node as visited and print it
    visited[v] = true;
    cout << v << " "; 
    // Recur for all the vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}

Graph Graph::getTranspose(){
    Graph g(V);
    for (int v = 0; v < V; v++){
        // Recur for all the vertices adjacent to this vertex
        list<int>::iterator i;
        for (i = adj[v].begin(); i != adj[v].end(); ++i){
            g.adj[*i].push_back(v);
        }
    }
    return g;
}

void Graph::addEdge(int v, int w){
    adj[v].push_back(w); // Add w to v’s list.
}

void Graph::fillOrder(int v, bool visited[], stack<int> &Stack){
    // Mark the current node as visited and print it
    visited[v] = true; 
    // Recur for all the vertices adjacent to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            fillOrder(*i, visited, Stack); 
    // All vertices reachable from v are processed by now, push v to Stack
    Stack.push(v);
}

// The main function that finds and prints all strongly connected components
int Graph::printSCCs(){
    stack<int> Stack;
    // Mark all the vertices as not visited (For first DFS)
    bool *visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
    // Fill vertices in stack according to their finishing times
    for (int i = 0; i < V; i++)
        if (visited[i] == false)
            fillOrder(i, visited, Stack);
    // Create a reversed graph
    Graph gr = getTranspose();
    // Mark all the vertices as not visited (For second DFS)
    for (int i = 0; i < V; i++)
        visited[i] = false;
    int count = 0;
    // Now process all vertices in order defined by Stack
    while (Stack.empty() == false){
    // Pop a vertex from stack
    int v = Stack.top();
    Stack.pop();
    // Print Strongly connected component of the popped vertex
    if (visited[v] == false){
        gr.DFSUtil(v, visited);
        cout << endl;
    }
    count++;
}
return count;
}
//----------------------------------------------------------------------------------------------- 
// test above functions
int main(){
    // Create a graph given in the above diagram
    Graph g(5);
    g.addEdge(1, 0);
    g.addEdge(0, 2);
    g.addEdge(2, 1);
    g.addEdge(0, 3);
    g.addEdge(3, 4);

    cout << "Following are strongly connected components in given graph \n";
    if (g.printSCCs() > 1){
        cout << "Graph is weakly connected.";
    }else{
        cout << "Graph is strongly connected.";
    }
    return 0;
}

如何找到图的最大连通分量？

1 个答案: