Assume that you are organizing a global tour with the goal of

seeing every nation. Kosaraju's Algorithm functions similarly

to a clever travel planner. It assists you in locating groupings

of nations where you can move easily between any two without

ever leaving the group. The program effectively investigates

links between nations, much like when you plan your trip,

which facilitates navigating the intricate web of international

travel routes.

// 1. Topological sort of original graph G

// 2. Itearte over all the vertices on reverse graph RG, and perform dfs

      on reverse graph 

// 3. Mark all the current visited node as one component

​

Note: its like onion-peeling algorithm and removal of sink nodes in 

Condensed connected graph

// =======================================================

​

#include <bits/stdc++.h>

using namespace std;

​

const int N = 1e5 + 1;

​

vector<int> gr[N], grr[N];

int vis[N];

int color[N];

vector<int> order;

​

void dfs1(int src) {

    vis[src] = 1;

​

    for (auto nbr : gr[src]) {

        if (!vis[nbr]) {

            dfs1(nbr);

        }

    }

​

    // i am at this point, bcz i have pushed all subtree

    order.push_back(src);

}

​

void dfs2(int src, int comp) {

    vis[src] = 1;

    color[src] = comp;

    for (auto nbr : grr[src]) {

        if (!vis[nbr]) {

            dfs2(nbr, comp);

        }

    }

​

}

​

int main() {

​

#ifndef ONLINE_JUDGE

    freopen("input.txt", "r", stdin);

    freopen("output.txt", "w", stdout);

#endif

​

    int n, m;

    cin >> n >> m;

​

    for (int i = 0; i < m; i++) {

        int x, y;

        cin >> x >> y;

        gr[x].push_back(y);

        grr[y].push_back(x);

    }

​

    for (int i = 1; i <= n; i++) {

        if(!vis[i]){

            dfs1(i);

        }

    }

​

    reverse(order.begin(), order.end());

    memset(vis, 0, sizeof(vis));

​

    int comp = 1;

​

    for (auto x : order) {

        if (!vis[x]) {

            dfs2(x, comp++);

        }

    }

​

    cout << "Total SCC are " << comp - 1 << endl;

​

    for(int i=1;i<=n;i++){

        cout<<i<<"-->"<<color[i]<<endl;

    }

​

    return 0;

}

​