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I find this solution to be the easiest to understand:

It employs the Inclusion-Exclusion Principle, but the main recursive function is much simpler to understand.

#include <bits/stdc++.h>

This is the official explanation for the problem: Indian National Olympiad in Informatics Btw, this problem was part of the 2004 INOI

Yeah, your solution seems a lot more intuitive, thanks for sharing. :)

Thank you so much for this excellent article and all the best for your future. I hope you will continue to write more like this.

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#include <bits/stdc++.h>
#include <list>
long long N;
std::vector <long long> graphs[100010];
bool vis[100010];

arr=list(map(int,input().split()))
n=arr[0]
graph=[[]for i in range(n+1)]
cost=[]
for i in range(1,n+1):

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The best and easiest way I have found is the Directed Graph implementation using STL vector class. I have recently researched about all of its methods to provide phd dissertation writing help ...

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Do we have to assume that the number of roads before and after the reform are same?

Wrong answer on test case 10.

Please, help me. Here's my code:

My logic is that the graph may be a collection of disconnected graphs. So a disconncted graph has a cycle=0 isolated cities, otherwise 1 isolated cities.

#include <bits/stdc++.h>
 
using namespace std;
long long n,m,x,y;
vector <long long> graph[1000000];
bool vis[1000000];
int main()
{
    long long cycle (long long,long long);
    cin>>n>>m;
 
    for(long long i = 0 ; i < m ; i++)
    {
        cin>>x>>y;
        graph[x].push_back(y);
        graph[y].push_back(x);
    }
 
 
    for(long long i = 1 ; i <= n ; i++ )       //This loops helps in navigating through different disconnected graphs
    {
         x=0;                                   // x is used inside cycle()
        if( !vis[ i ] ){   
            vis[1]=1;
            isolated += ( !cycle( i , -1 ) ) ; //if it is a cycle add 0, else add 1
        }
    }
 
    cout<<isolated;
 
    return 0;
}
 
long long cycle(long long cur, long long prev)
{
    long long x=0;
 
    for(unsigned long long i = 0 ; i < graph[cur].size(); i++ )
    { 
        if( !vis[ graph[ cur ][ i ] ] )
        {
            vis[ graph[ cur ][ i ] ] = 1;
            x+=cycle( graph[ cur ][ i ] , cur );
        }
        else if( prev != graph[ cur ][ i ] )
             {
                 x++;   //It has a cycle (so we will add 0 to isolated.
             }
    }
    return x;
}
 

This might help: International Olympiad in Informatics: Path to Gold

This playlist is mostly based on IARCS syllabus and previous years' papers' trends.