SPOJ Solution-Longest path in a tree-Solution in C, C++, Java, Python

Input:
3
1 2
2 3

Output:
2

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h7gt;
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
 
#define MOD (ll)1000000007
#define pb   push_back
#define EPS 1e-9
#define FOR(i,n)  for(int i = 0;i < n; i++)
#define FORE(i,a,b) for(int i = a;i <= b; i++)
#define pi(a)   printf("%d\n", a)
#define all(c)  c.begin(), c.end()
#define tr(container, it)   for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
#define sdi(a, b)   si(a);si(b)
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl '\n'
#define F first
#define S second
#define FILL(arr, val)  memset(arr, val, sizeof(arr))
#define read(arr, n)    for(int i = 0;i  <  n; i++)cin>>arr[i];
#define sp ' '

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

void si(int &x){
    register int c = gc();
    x = 0;
    int neg = 0;
    for(;((c<48 || c>57) && c != '-');c = gc());
    if(c=='-') {neg=1;c=gc();}
    for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
    if(neg) x=-x;
}

const int MAXN = 10005;
vector<int> adj[MAXN];
bool vis[MAXN];
int currentMax;
int vertofound;


void dfs(int source, int maxlen){
    vis[source] = 1;
    if(maxlen > currentMax){
        currentMax = maxlen;
        vertofound = source;
    }
    for(int i = 0;i < adj[source].size(); i++){
        if(!vis[adj[source][i]])
            dfs(adj[source][i], maxlen+1);
    }
}

int main(){
    io;
    currentMax = INT_MIN;
    int nodes;
    cin >> nodes;
    for(int i = 1;i <= nodes-1; i++){
        int a, b;
        cin >> a >> b;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    dfs(1, 0);
    for(int i = 0;i <= nodes; i++)
        vis[i] = 0;
    dfs(vertofound, 0);
    cout << currentMax << endl;
    return 0;
}

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Input

Output