HackerEarth Longest Path problem solution

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define be begin()
#define en end()
#define le length()
#define sz size()
#define all(x) (x).begin(),(x).end()
#define alli(a, n, k) (a+k),(a+n+k)
#define REP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)
#define REPI(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)
#define REPITER(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)
#define REPEACH(i, a) for(auto& i:a)

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd
#define have adsgagshdshfhds

#define eps 1e-6
#define pi 3.141592653589793

using namespace std;

template<class T> inline T gcd(T a, T b) { while(b) b ^= a ^= b ^= a %= b; return a; }
template<class T> inline T mod(T x) { if(x < 0) return -x; else return x; }

typedef vector<int> VII;
typedef vector<ll> VLL;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<int, PII > PPII;
typedef vector< PII > VPII;
typedef vector< PPII > VPPI;

const int MOD = 1e9 + 7;
const int INF = 1e9;

// Template End
const int MAX = 1e5 + 5;
int a[MAX];
int ans = 0;
VII adj[MAX];


int dfs(int x, int par) {
    int mx1 = 0, mx2 = 0;
    REPEACH(y, adj[x]) {
        if (y != par) {
            int p = dfs(y, x);
            if (p > mx1) mx2 = mx1, mx1 = p;
            else if (p > mx2) mx2 = p;
        }
    }
    if (a[x] == 0) {
        ans = max(ans, mx1 + 1);
        return mx1 + 1;
    }
    else return 0;
}

int main(int argc, char* argv[]) {
    if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);
    if(argc == 3) freopen(argv[2], "w", stdout);
    ios::sync_with_stdio(false);
    int n, k, u, v;
    cin >> n >> k;
    REP(i, 1, n+1, 1) {
        cin >> a[i];
        a[i] = (a[i] % k) ? 1 : 0;
    }
    REP(i, 1, n, 1) {
        cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    ans = max(ans, dfs(1, -1));
    cout << ans-1 << endl;
    return 0;
}