Finding number of connected components in an undirected graph

Question

The users inputs the number of vertices ( n ) and then - in the next n lines - how the vertices are connected, namely that the number x in i -th line means that vertex i is connected with vertex x (the graph is undirected). The task is to find the number of connected components in this graph, and my code - for some reason I'm not able to find - outputs wrong values (for example, with input 4 2 1 2 4 , my code outputs 4 instead of 2 ). Any help greatly appreciated.

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

int n;

using namespace std;
vector <int> graph[1000006];
int components[1000006];
int no_of_components;
stack <int> mystack;

int main(){

    cin >> n;

    for (int i=0; i<n; i++){
        int X;
        cin >> X;
        graph[X-1].push_back(i);
    }

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

        if (components[i]>0) continue;

        no_of_components++;

        mystack.push(i);
        components[i]=no_of_components;

        while (mystack.empty()==false){
            int v;

            v=mystack.top();
            mystack.pop();

            for (int u=0; u<graph[v].size(); u++){
                if (components[u]>0) continue;

                mystack.push(u);
                components[u]=no_of_components;

            }
        }
    }

    cout << no_of_components;

    return 0;

}

Answer 1

In your code there is a confusion between the counter u which allows you to iterate through the connections of the node v, and the connections themselves:

• v is a node
• graph[v] is the vector of connections
• u is an index in in the vector of connections
• so graph[v][u] is the node connected to v and component[graph[v][u]] the component marker you have to update

So you have to correct the inner for-loop as follows:

        for (int u=0; u<graph[v].size(); u++){
            if (components[graph[v][u]]>0) continue;

            mystack.push(graph[v][u]);
            components[graph[v][u]]=no_of_components;

        }

Then works as expected.

Online demo