HackerEarth Graphs problem solution

In this HackerEarth Graphs problem solution, You are given an undirected graph G that contains n nodes and m edges. It is also mentioned that G does not contain any cycles. A sequence of nodes (A1,A2,A3,...Ak) is special if distance  d(Ai.Ai+1) = f.i for all 1<=i<k , and all Ai are distinct. Determine the size of a maximal special sequence. Hence, the one with maximum k.

HackerEarth Graphs problem solution.

#include <bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define memreset(a) memset(a,0,sizeof(a))
#define testcase(t) int t;cin>>t;while(t--)
#define forstl(i,v) for(auto &i: v)
#define forn(i,e) for(int i = 0; i < e; i++)
#define forsn(i,s,e) for(int i = s; i < e; i++)
#define rforn(i,s) for(int i = s; i >= 0; i--)
#define rforsn(i,s,e) for(int i = s; i >= e; i--)
#define leadzero(a) __builtin_clz(a) // count leading zeroes
#define trailzero(a) __builtin_ctz(a) // count trailing zeroes
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln endl
#define dbg(x) cerr<<#x<<" = "<<x<<endl
#define dbg2(x,y) cerr<<#x<<" = "<<x<<" & "<<#y<<" = "<<y<<endl;
#define dbgstl32(v) cerr<<#v<<" = \n"; { int c=0; forstl(it,v) \
cerr<<"    Term "<< ++c <<" = "<<it<<ln;} cerr<<endl
#define dbgstlp32(v) cerr<<#v<<" = \n"; { int c=0; forstl(it,v) \
cerr<<"    Term "<< ++c <<" = "<<it.fi<<" , "<<it.se<<ln;} cerr<<endl
#define dbgarr(v,s,e) cerr<<#v<<" = "; forsn(i,s,e) cerr<<v[i]<<", "; cerr<<endl
#define inp freopen("input.txt", "r", stdin)
#define outp freopen("output.txt", "w", stdout)
#define dbg(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); \
stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) { cerr<<endl; }
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << "\t"; err(++it, args...);
}
template<typename T1,typename T2>
ostream& operator <<(ostream& c,pair<T1,T2> &v){
    c<<"("<<v.fi<<","<<v.se<<")"; return c;
}
template <template <class...> class TT, class ...T>
ostream& operator<<(ostream& out,TT<T...>& c){
    out<<"{ ";
    forstl(x,c) out<<x<<" ";
    out<<"}"; return out;
}
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<int> v32; 
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;
const ll MOD = 1e9+7, LIM = 1e5+5;
const ld EPS = 1e-9;

int main(){
    fastio;
        
    int n,m,f;
    cin>>n>>m>>f;
    v32 adj[n];
    forn(i,m){
        int x,y;
        cin>>x>>y;
        x--;
        y--;
        adj[x].pb(y);
        adj[y].pb(x);
    }

    int visited[n] = {0};
    int par[n] = {0};
    int dist[n] = {0};

    int maxval = 0, posn = -1;

    forn(i,n){
        if(visited[i]>0) continue;

        queue<int> q;
        visited[i] = 1;
        q.push(i);
        int t = 0;
        while(!q.empty()){
            t = q.front();
            q.pop();

            forstl(r, adj[t]){
                if(visited[r]>0) continue;
                visited[r] = 1;
                q.push(r);
            }
        }
        int end = t;

        q.push(end);
        visited[end] = 2;
        dist[end] = 1;
        par[end] = -1;
        while(!q.empty()){
            t = q.front();
            q.pop();

            forstl(r, adj[t]){
                if(visited[r]>1) continue;
                visited[r] = 2;
                q.push(r);
                dist[r] = dist[t]+1;
                par[r] = t;
            }
        }
        if(dist[t]>maxval){
            maxval = dist[t];
            posn = t;
        }
    }

    v32 v;
    int num = 0;
    while(posn != -1){
        if(num == 0) v.pb(posn);
        num++;
        if(num == f) num = 0;
        posn = par[posn];
    }

    v32 answer;
    int g = v.size()/2;
    int ptr1 = g, ptr2 = g-1;

    while(ptr1 <v.size() || ptr2 >=0){
        answer.pb(v[ptr1]);
        if(ptr2 < 0) break;
        answer.pb(v[ptr2]); 
        ptr1++;
        ptr2--;
    }

    cout<<answer.size()<<ln;

    return 0;
}