Problem p02573 (C++) — buggy submission, verdict: Time Limit Exceeded

/*
AtCoder Beginner Contest 177- D - Friends
https://atcoder.jp/contests/abc177/tasks/abc177_d
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005ll;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x)                                                                \\
  for (auto &(z) : x)                                                          \\
    cin >> z;
#define FOR(a, b, c) for (int(a) = (b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a) = (b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a) = (b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a) = (b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a) = (b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto &(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP                                                               \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL)

class dsu {
private:
  vector<ll> d, sz;

public:
  dsu(ll n) {
    d = vector<ll>(n);
    sz = vector<ll>(n, 1);
    REP(i, n) d[i] = i;
  }

  ll p(ll a) {
    if (d[a] == a)
      return a;
    else
      return d[a] = p(d[a]);
  }

  ll get_sz(ll a) { return sz[p(a)]; }

  void unite(ll a, ll b) {
    ll pa = p(a);
    ll pb = p(b);
    if (pa == pb)
      return;
    else {
      sz[pb] += sz[pa];
      sz[pa] = 0;
      d[pa] = pb;
      return;
    }
    return;
  }
};

// Solution 1: Use DSU template to find the max tree size
void solve() {
  ll n, m, a, b;
  cin >> n >> m;
  dsu d = dsu(n);
  REP(i, m) {
    cin >> a >> b;
    --a, --b;
    d.unite(a, b);
  }
  int ans = -INT_MAX;
  REP(i, n) MAX(ans, (int)d.get_sz(i));
  OUT(ans);
}

void solve2() {
  ll n, m, a, b;
  cin >> n >> m;
  vvi g(n);
  vi vis(n, 0);
  REP(i, m) {
    cin >> a >> b;
    --a, --b;
    g[a].pb(b);
    g[b].pb(a);
  }
  int ans = 0, cnt;
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (vis[i])
      continue;
    vis[i] = 1, cnt = 0;
    q.push(i);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      cnt++;
      EACH(v, g[u]) {
        if (!vis[v]) {
          q.push(v);
          vis[v] = 1;
        }
      }
    }
    MAX(ans, cnt);
  }
  OUT(ans);
}

int main() {
  FAST_INP;
  //    #ifndef ONLINE_JUDGE
  //    freopen("input.txt","r", stdin);
  //    freopen("output.txt","w", stdout);
  //    #endif
  //
  //    int tc; cin >> tc;
  //    TC(tc) solve();
  solve2();
  return 0;
}


Error: TLE

Fixed submission:

/*
AtCoder Beginner Contest 177- D - Friends
https://atcoder.jp/contests/abc177/tasks/abc177_d
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005ll;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
const ll MOD = 1000000007;

#define DEBUG fprintf(stderr, "====TESTING====\\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x)                                                                \\
  for (auto &(z) : x)                                                          \\
    cin >> z;
#define FOR(a, b, c) for (int(a) = (b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a) = (b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a) = (b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a) = (b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a) = (b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto &(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP                                                               \\
  ios_base::sync_with_stdio(false);                                            \\
  cin.tie(NULL)

class dsu {
private:
  vector<ll> d, sz;

public:
  dsu(ll n) {
    d = vector<ll>(n);
    sz = vector<ll>(n, 1);
    REP(i, n) d[i] = i;
  }

  ll p(ll a) {
    if (d[a] == a)
      return a;
    else
      return d[a] = p(d[a]);
  }

  ll get_sz(ll a) { return sz[p(a)]; }

  void unite(ll a, ll b) {
    ll pa = p(a);
    ll pb = p(b);
    if (pa == pb)
      return;
    else {
      sz[pb] += sz[pa];
      sz[pa] = 0;
      d[pa] = pb;
      return;
    }
    return;
  }
};

// Solution 1: Use DSU template to find the max tree size
void solve() {
  ll n, m, a, b;
  cin >> n >> m;
  dsu d = dsu(n);
  REP(i, m) {
    cin >> a >> b;
    --a, --b;
    d.unite(a, b);
  }
  int ans = -INT_MAX;
  REP(i, n) MAX(ans, (int)d.get_sz(i));
  OUT(ans);
}

void solve2() {
  ll n, m, a, b;
  cin >> n >> m;
  vvi g(n);
  vi vis(n, 0);
  REP(i, m) {
    cin >> a >> b;
    --a, --b;
    g[a].pb(b);
    g[b].pb(a);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i])
      continue;
    vis[i] = 1;
    int cnt = 0;
    queue<int> q;
    q.push(i);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      cnt++;
      EACH(v, g[u]) {
        if (!vis[v]) {
          q.push(v);
        }
      }
    }
    MAX(ans, cnt);
  }
  OUT(ans);
}

int main() {
  FAST_INP;
  //    #ifndef ONLINE_JUDGE
  //    freopen("input.txt","r", stdin);
  //    freopen("output.txt","w", stdout);
  //    #endif
  //
  //    int tc; cin >> tc;
  //    TC(tc) solve();
  solve2();
  return 0;
}