why does kosaraju algorithm works and what is the idea behind it and is this a correct implementation?

Question

why do we create a transpose of the graph and then run dfs on the transpose in the second pass.I've tried reading proof of correctness http://www.jeffreykarres.com/blog/kosaraju/ online but couldn't understand is there some alternative approach to do this which is easy to understand

here is my implementation of the algorithm it takes number of vetices and edges as inputs and then takes directed edges as inputs(vertices are numbered 0 to n-1)

#include <bits/stdc++.h>
using namespace std;
void dfsForward(int src , vector<vector<int>> g , vector<bool> &vis, stack<int> &s ){
    vis[src]=true;
    for(int i=0;i<g[src].size();i++){
        if(!vis[g[src][i]])
         dfsForward(g[src][i],g,vis,s);
    }
    s.push(src);
}
void dfsRev(int src , vector<vector<int>> t , vector<bool> &vis, vector<int> &comp,int count ){
    vis[src]=true;
    for(int i=0;i<t[src].size();i++){
        if(!vis[t[src][i]]){
           comp.push_back(t[src][i]);
            dfsRev(t[src][i],t,vis,comp,count);
        }
    }
}
vector<vector<int>> kosaraju(vector<vector<int>> g,vector<vector<int>> t, int n){
    vector<bool> vis(n,false); 
    vector<bool> visRev(n,false); 
    vector<vector<int>> scc; 
    int count=0;
    stack<int> s;
    for(int i=0;i<n;i++){
        if(!vis[i])
         dfsForward(i,g,vis,s);
    }
    while(!s.empty()){
        int temp =s.top();
        s.pop();
        if(!visRev[temp]){
           count++;    
           vector<int> comp;
           comp.push_back(temp);
           dfsRev(temp,t,visRev,comp,count);
           scc.push_back(comp);
        }
    } 
    return scc;
}
int main() {
    int n,e,u,v;
    cin>>n>>e;
    vector<vector<int>> g(n);
    vector<vector<int>> t(n);
    for(int i=0;i<e;i++){
        cin>>u>>v;
        g[u].push_back(v);
        t[v].push_back(u);
    }
    cout<<"components are "<<endl;
    vector<vector<int>> scc = kosaraju(g,t,n);
    for(int i=0;i<scc.size();i++){
        vector<int> arr = scc[i];
        for(int j=0;j<arr.size();j++){ 
            cout<<arr[j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}

Answer 1

Kosaraju algorithm works in three simple steps:

1. Reverse the Graph G to form G'.
2. Apply DFS-Loop in G' and calculate ordering of each vertex (we call it finishing times).
3. Apply DFS-Loop in G in descending order of finishing times.

To answer your question: Why we reverse the graph in first step?

As you know, SCCs in Graph G and reverse graph G' will always be the same. Consider each SCC a separate node N, which defines the "graphs of SCCs" S:=G and S':= G', then both S and S' will be DAGs (directed acyclic graphs).

(*) As a property of DFS on a DAG, the largest finishing time will always be a sink vertex (a vertex from which there is no outgoing arc).

When we ran DFS on G' terminating at v, we have effectively also run DFS on S' terminating at N, and we must have v being a member of N. But (*) states that N is a sink vertex. So if you re-run DFS on G from v, it will only iterate through N. Hence this way iteratively you will uncover all the SCCs in the graph.