kosaraju.cpp

/* Implementation of Kosaraju's Algorithm to find out the strongly connected
   components (SCCs) in a graph. Author:Anirban166
*/

#include <iostream>
#include <stack>
#include <vector>

/**
 * Iterative function/method to print graph:
 * @param a adjacency list representation of the graph
 * @param V number of vertices
 * @return void
 **/
void print(const std::vector<std::vector<int> > &a, int V) {
    for (int i = 0; i < V; i++) {
        if (!a[i].empty()) {
            std::cout << "i=" << i << "-->";
        }
        for (int j : a[i]) {
            std::cout << j << " ";
        }
        if (!a[i].empty()) {
            std::cout << std::endl;
        }
    }
}

/**
 * //Recursive function/method to push vertices into stack passed as parameter:
 * @param v vertices
 * @param st stack passed by reference
 * @param vis array to keep track of visited nodes (boolean type)
 * @param adj adjacency list representation of the graph
 * @return void
 **/
void push_vertex(int v, std::stack<int> *st, std::vector<bool> *vis,
                 const std::vector<std::vector<int> > &adj) {
    (*vis)[v] = true;
    for (auto i = adj[v].begin(); i != adj[v].end(); i++) {
        if ((*vis)[*i] == false) {
            push_vertex(*i, st, vis, adj);
        }
    }
    st->push(v);
}

/**
 * //Recursive function/method to implement depth first traversal(dfs):
 * @param v vertices
 * @param vis array to keep track of visited nodes (boolean type)
 * @param grev graph with reversed edges
 * @return void
 **/
void dfs(int v, std::vector<bool> *vis,
         const std::vector<std::vector<int> > &grev) {
    (*vis)[v] = true;
    // cout<<v<<" ";
    for (auto i = grev[v].begin(); i != grev[v].end(); i++) {
        if ((*vis)[*i] == false) {
            dfs(*i, vis, grev);
        }
    }
}

// function/method to implement Kosaraju's Algorithm:
/**
* Info about the method
* @param V vertices in graph
* @param adj array of vectors that represent a graph (adjacency list/array)
* @return int ( 0, 1, 2..and so on, only unsigned values as either there can be
no SCCs i.e. none(0) or there will be x no. of SCCs (x>0)) i.e. it returns the
count of (number of) strongly connected components (SCCs) in the graph.
(variable 'count_scc' within function)
**/
int kosaraju(int V, const std::vector<std::vector<int> > &adj) {
    std::vector<bool> vis(V, false);
    std::stack<int> st;
    for (int v = 0; v < V; v++) {
        if (vis[v] == false) {
            push_vertex(v, &st, &vis, adj);
        }
    }
    // making new graph (grev) with reverse edges as in adj[]:
    std::vector<std::vector<int> > grev(V);
    for (int i = 0; i < V + 1; i++) {
        for (auto j = adj[i].begin(); j != adj[i].end(); j++) {
            grev[*j].push_back(i);
        }
    }
    // cout<<"grev="<<endl; ->debug statement
    // print(grev,V);       ->debug statement
    // reinitialise visited to 0
    for (int i = 0; i < V; i++) vis[i] = false;
    int count_scc = 0;
    while (!st.empty()) {
        int t = st.top();
        st.pop();
        if (vis[t] == false) {
            dfs(t, &vis, grev);
            count_scc++;
        }
    }
    // cout<<"count_scc="<<count_scc<<endl; //in case you want to print here
    // itself, uncomment & change return type of function to void.
    return count_scc;
}

// All critical/corner cases have been taken care of.
// Input your required values: (not hardcoded)
int main() {
    int t = 0;
    std::cin >> t;
    while (t--) {
        int a = 0, b = 0;  // a->number of nodes, b->directed edges.
        std::cin >> a >> b;
        int m = 0, n = 0;
        std::vector<std::vector<int> > adj(a + 1);
        for (int i = 0; i < b; i++)  // take total b inputs of 2 vertices each
                                     // required to form an edge.
        {
            std::cin >> m >> n;  // take input m,n denoting edge from m->n.
            adj[m].push_back(n);
        }
        // pass number of nodes and adjacency array as parameters to function:
        std::cout << kosaraju(a, adj) << std::endl;
    }
    return 0;
}