Problem p02678 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <algorithm>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <vector>

using namespace std;

// マクロ&定数&関数 ================================================
typedef unsigned int uint;
typedef long long ll;

typedef pair<ll, ll> pll;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vpint> vvpint;
typedef vector<vpll> vvpll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<string>> vvstring;
typedef vector<vector<bool>> vvbool;

typedef vector<vector<vector<ll>>> vvvll;

const int c_YET = -1;
const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = {0, 0, 1, -1, 1, 1, -1, -1, 0}; // 4;４近傍
const int DY[9] = {1, -1, 0, 0, 1, -1, 1, -1, 0}; // 8:８近傍 9:(0,0)を含む
const ll MOD = 998244353;                         // 10^9 + 7
const ll MAX = 1e9;
const double PI = 3.14159265358979323846264338327950288;

//---------------------------------------------------------------
//  オーバーフローチェック
//---------------------------------------------------------------
bool is_overflow(ll a, ll b) { return ((a * b) / b != a); }

//---------------------------------------------------------------
//  約数列挙
//---------------------------------------------------------------
vll divisor(ll n) {
  vll ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n)
        ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

//---------------------------------------------------------------
// N以下のすべての素数を列挙する(エラトステネスの篩)
//---------------------------------------------------------------
vbool searchSosuu(ll N) {
  vbool sosuu; // ←これをグローバル変数にして１度だけ実行する
  for (ll i = 0; i < N; i++) {
    sosuu.emplace_back(true);
  }
  sosuu[0] = false;
  sosuu[1] = false;
  for (ll i = 2; i < N; i++) {
    if (sosuu[i]) {
      for (ll j = 2; i * j < N; j++) {
        sosuu[i * j] = false;
      }
    }
  }

  return sosuu;
}

//---------------------------------------------------------------
// 素因数分解　O(√N)
//---------------------------------------------------------------
vpll div_prime(ll n) {
  vpll prime_factor;
  for (ll i = 2; i * i <= n; i++) {
    ll count = 0;
    while (n % i == 0) {
      count++;
      n /= i;
    }
    if (count) {
      pair<ll, ll> temp = {i, count};
      prime_factor.emplace_back(temp);
    }
  }
  if (n != 1) {
    pair<ll, ll> temp = {n, 1};
    prime_factor.emplace_back(temp);
  }

  return prime_factor;
}

//---------------------------------------------------------------
// 素数判定
//---------------------------------------------------------------
bool is_sosuu(ll N) {

  if (N < 2) {
    return false;
  } else if (N == 2) {
    return true;
  } else if (N % 2 == 0) {
    return false;
  }

  for (ll i = 3; i <= sqrt(N); i += 2) {
    if (N % i == 0) {
      return false;
    }
  }

  return true;
}

//---------------------------------------------------------------
// 最大公約数（ユークリッドの互除法）
//---------------------------------------------------------------
ll gcd(ll a, ll b) {
  if (a < b) {
    ll tmp = a;
    a = b;
    b = tmp;
  }

  ll r = a % b;
  while (r != 0) {
    a = b;
    b = r;
    r = a % b;
  }

  return b;
}

//---------------------------------------------------------------
// 最小公倍数
//---------------------------------------------------------------
ll lcm(ll a, ll b) {
  ll temp = gcd(a, b);
  return temp * (a / temp) * (b / temp);
}

//---------------------------------------------------------------
// 階乗
//---------------------------------------------------------------
ll factorial(ll n) {
  if (n <= 1) {
    return 1;
  }

  return (n * (factorial(n - 1))) % MOD;
}

//---------------------------------------------------------------
// 高速コンビネーション計算(前処理:O(N) 計算:O(1))
//---------------------------------------------------------------
// テーブルを作る前処理
ll comb_const = 200005;
vll fac(comb_const), finv(comb_const), inv(comb_const);
bool COMineted = false;
void COMinit() {

  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (ll i = 2; i < comb_const; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }

  COMineted = true;
}

// 二項係数計算
ll COM(ll n, ll k) {
  if (COMineted == false)
    COMinit();

  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//---------------------------------------------------------------
// 繰り返し２乗法 base^sisuu
//---------------------------------------------------------------
ll RepeatSquaring(ll base, ll sisuu) {
  if (sisuu < 0) {
    cout << "RepeatSquaring: 指数が負です!" << endl;
    return 0;
  }
  if (sisuu == 0)
    return 1;
  if (sisuu % 2 == 0) {
    ll t = RepeatSquaring(base, sisuu / 2) % MOD;
    return (t * t) % MOD;
  }
  return base * RepeatSquaring(base, sisuu - 1) % MOD;
}

//---------------------------------------------------------------
// 高速単発コンビネーション計算(O(logN))
//---------------------------------------------------------------
ll fast_com(ll a, ll b) {

  ll bunshi = 1;
  ll bunbo = 1;
  for (ll i = 1; i <= b; i++) {
    bunbo *= i;
    bunbo %= MOD;

    bunshi *= (a - i + 1);
    bunshi %= MOD;
  }

  ll ret = bunshi * RepeatSquaring(bunbo, MOD - 2);

  ret %= MOD;
  while (ret < 0) {
    ret += MOD;
  }

  return ret;
}

//---------------------------------------------------------------
// 2直線の交差判定(直線(x1, y1)->(x2, y2) と 直線(X1, Y1)->(X2, Y2))
//---------------------------------------------------------------
bool is_cross(ll x1, ll y1, ll x2, ll y2, ll X1, ll Y1, ll X2, ll Y2) {
  ll dx_ai = X1 - x1;
  ll dy_ai = Y1 - y1;
  ll dx_bi = X1 - x2;
  ll dy_bi = Y1 - y2;

  ll dx_ai2 = X2 - x1;
  ll dy_ai2 = Y2 - y1;
  ll dx_bi2 = X2 - x2;
  ll dy_bi2 = Y2 - y2;

  ll si = dx_ai * dy_bi - dy_ai * dx_bi;
  ll si2 = dx_ai2 * dy_bi2 - dy_ai2 * dx_bi2;

  ll si3 = dx_ai * dy_ai2 - dy_ai * dx_ai2;
  ll si4 = dx_bi * dy_bi2 - dy_bi * dx_bi2;

  return (si * si2 < 0 && si3 * si4 < 0);
}

//---------------------------------------------------------------
// 最長増加部分列の長さ(O(NlogN))
//---------------------------------------------------------------
ll LSI(vll vec, ll size) {
  vll lsi(size + 1); // 長さjを作った時の右端の最小値
  for (ll i = 0; i <= size; i++) {
    lsi[i] = LLINF;
  }

  lsi[0] = 0;
  lsi[1] = vec[0];

  for (ll i = 1; i < size; i++) {
    // 初めてvec[i]の方が小さくなるところを探す
    auto Iter = lower_bound(lsi.begin(), lsi.end(), vec[i]);
    ll idx = Iter - lsi.begin();

    if (idx > 0 && lsi[idx - 1] < vec[i]) // 条件文の前半怪しい
    {
      lsi[idx] = vec[i];
    }
  }

  for (ll i = size; i >= 0; i--) {
    if (lsi[i] < LLINF) {
      return i;
    }
  }
}

//---------------------------------------------------------------
// 木の根からの深さ
//---------------------------------------------------------------
vll tree_depth(vvll edge, ll start_node, ll n_node) {
  vll dist(n_node, LLINF);
  dist[start_node] = 0;

  stack<pll> S;
  S.push({start_node, 0});

  while (!S.empty()) {
    ll node = S.top().first;
    ll d = S.top().second;
    dist[node] = d;

    S.pop();

    for (int i = 0; i < edge[node].size(); i++) {
      if (dist[edge[node][i]] == LLINF) {
        S.push({edge[node][i], d + 1});
      }
    }
  }

  return dist;
}

//---------------------------------------------------------------
// ワーシャルフロイド法(O(N^3)) 任意の2点間の最短距離
//---------------------------------------------------------------
vvll warshall_floyd(ll n, vvll d) {
  for (int k = 0; k < n; k++) {     // 経由する頂点
    for (int i = 0; i < n; i++) {   // 始点
      for (int j = 0; j < n; j++) { // 終点
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  return d;
}

//---------------------------------------------------------------
// Union Find
//---------------------------------------------------------------
/*
        UnionFind uf(要素の個数);
        for(int i = 0;i < 関係の個数; i++)
        {
                uf.merge(A[i], B[i]);
        }

        nを含む集合の大きさ = uf.size(n)
        nを含む集合の代表者 = uf.root(n)
        集合の個数 = uf.n_group
*/
class UnionFind {
public:
  vector<ll> par; // 各元の親を表す配列
  vector<ll> siz; // 素集合のサイズを表す配列(1 で初期化)
  ll n_group;     // 集合の数

  // Constructor
  UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
    for (ll i = 0; i < sz_; ++i)
      par[i] = i; // 初期では親は自分自身
    n_group = sz_;
  }
  void init(ll sz_) {
    par.resize(sz_);
    siz.assign(sz_, 1LL); // resize だとなぜか初期化されなかった
    for (ll i = 0; i < sz_; ++i)
      par[i] = i; // 初期では親は自分自身
  }

  // Member Function
  // Find
  ll root(ll x) { // 根の検索
    while (par[x] != x) {
      x = par[x] = par[par[x]]; // x の親の親を x の親とする
    }
    return x;
  }

  // Union(Unite, Merge)
  bool merge(ll x, ll y) {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    // merge technique（データ構造をマージするテク．小を大にくっつける）
    if (siz[x] < siz[y])
      swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    n_group--;
    return true;
  }

  bool issame(ll x, ll y) { // 連結判定
    return root(x) == root(y);
  }

  ll size(ll x) { // 素集合のサイズ
    return siz[root(x)];
  }
};

//========================================================================

struct edge {
  int to, cost;
};
typedef vector<vector<edge>> AdjList;
AdjList graph;

typedef pair<int, int> P;

vector<int> dist;
vector<int> prever;

void dijkstra(int n, int s) {
  dist = vector<int>(n, INF);
  prever = vector<int>(n, -1);
  dist[s] = 0;

  priority_queue<P, vector<P>, greater<P>> que;
  que.push(P(0, s));

  while (!que.empty()) {
    P p = que.top();
    que.pop();
    int v = p.second;
    if (dist[v] < p.first) {
      continue;
    }
    for (int i = 0; i < graph[v].size(); i++) {
      edge e = graph[v][i];
      if (dist[e.to] > dist[v] + e.cost) {
        dist[e.to] = dist[v] + e.cost;
        prever[e.to] = v;
        que.push(P(dist[e.to], e.to));
      }
    }
  }
}

int main() {

  ////==================================
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(30);
  ////==================================

  int n, m;
  cin >> n >> m;

  vvll near(n);

  graph = AdjList(n);
  for (int i = 0; i < m; i++) {
    int from, to, cost;
    cin >> from >> to;
    from--;
    to--;
    edge t = {to, 1};
    edge t2 = {from, 1};
    graph[from].push_back(t);
    graph[to].push_back(t2);

    near[from].emplace_back(to);
    near[to].emplace_back(from);
  }

  dijkstra(n, 0);

  vll ans(n, -1);
  for (int i = 1; i < n; i++) {
    ll d = dist[i];

    // d-1で連結
    for (int j = 0; j < near[i].size(); j++) {
      if (dist[near[i][j]] == d - 1) {
        ans[i] = near[i][j];
      }
    }

    if (ans[i] == -1) {
      cout << "No";
      return 0;
    }
  }

  cout << "Yes" << endl;
  for (int i = 1; i < n; i++) {
    cout << ans[i] + 1 << endl;
  }
}


Error: TLE

Fixed submission:

#include <algorithm>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <vector>

using namespace std;

// マクロ&定数&関数 ================================================
typedef unsigned int uint;
typedef long long ll;

typedef pair<ll, ll> pll;

typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<bool> vbool;
typedef vector<string> vstring;

typedef vector<pair<int, int>> vpint;
typedef vector<pair<ll, ll>> vpll;
typedef vector<pair<double, double>> vpdouble;

typedef vector<vector<int>> vvint;
typedef vector<vector<ll>> vvll;
typedef vector<vpint> vvpint;
typedef vector<vpll> vvpll;
typedef vector<vector<double>> vvdouble;
typedef vector<vector<string>> vvstring;
typedef vector<vector<bool>> vvbool;

typedef vector<map<ll, ll>> vmll;

typedef vector<vector<vector<ll>>> vvvll;

const int INF = 1e9 + 1;
const ll LLINF = 1e17 + 1;
const int DX[9] = {0, 0, 1, -1, 1, 1, -1, -1, 0}; // 4;４近傍
const int DY[9] = {1, -1, 0, 0, 1, -1, 1, -1, 0}; // 8:８近傍 9:(0,0)を含む
const ll MOD = 1e9 + 7;                           // 10^9 + 7
const ll MAX = 1e9;
const double pllI = 3.14159265358979323846264338327950288;

//---------------------------------------------------------------
//  オーバーフローチェック
//---------------------------------------------------------------
bool is_overflow(ll a, ll b) { return ((a * b) / b != a); }

//---------------------------------------------------------------
//  約数列挙
//---------------------------------------------------------------
vll divisor(ll n) {
  vll ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n)
        ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

//---------------------------------------------------------------
// N以下のすべての素数を列挙する(エラトステネスの篩)
//---------------------------------------------------------------
vbool searchSosuu(ll N) {
  vbool sosuu; // ←これをグローバル変数にして１度だけ実行する
  for (ll i = 0; i < N; i++) {
    sosuu.emplace_back(true);
  }
  sosuu[0] = false;
  sosuu[1] = false;
  for (ll i = 2; i < N; i++) {
    if (sosuu[i]) {
      for (ll j = 2; i * j < N; j++) {
        sosuu[i * j] = false;
      }
    }
  }

  return sosuu;
}

//---------------------------------------------------------------
// 素因数分解　O(√N)
//---------------------------------------------------------------
vpll div_prime(ll n) {
  vpll prime_factor;
  for (ll i = 2; i * i <= n; i++) {
    ll count = 0;
    while (n % i == 0) {
      count++;
      n /= i;
    }
    if (count) {
      pair<ll, ll> temp = {i, count};
      prime_factor.emplace_back(temp);
    }
  }
  if (n != 1) {
    pair<ll, ll> temp = {n, 1};
    prime_factor.emplace_back(temp);
  }

  return prime_factor;
}

//---------------------------------------------------------------
// 素数判定
//---------------------------------------------------------------
bool is_sosuu(ll N) {

  if (N < 2) {
    return false;
  } else if (N == 2) {
    return true;
  } else if (N % 2 == 0) {
    return false;
  }

  for (ll i = 3; i <= sqrt(N); i += 2) {
    if (N % i == 0) {
      return false;
    }
  }

  return true;
}

//---------------------------------------------------------------
// 最大公約数（ユークリッドの互除法）(O(logA))
//---------------------------------------------------------------
ll gcd(ll a, ll b) {
  if (a < b) {
    ll tmp = a;
    a = b;
    b = tmp;
  }

  ll r = a % b;
  while (r != 0) {
    a = b;
    b = r;
    r = a % b;
  }

  return b;
}

//---------------------------------------------------------------
// 最小公倍数
//---------------------------------------------------------------
ll lcm(ll a, ll b) {
  ll temp = gcd(a, b);
  return temp * (a / temp) * (b / temp);
}

//---------------------------------------------------------------
// 階乗
//---------------------------------------------------------------
ll factorial(ll n) {
  if (n <= 1) {
    return 1;
  }

  return (n * (factorial(n - 1))) % MOD;
}

//---------------------------------------------------------------
// 高速コンビネーション計算(前処理:O(N) 計算:O(1))
//---------------------------------------------------------------
// テーブルを作る前処理
ll comb_const = 200005;
vll fac(comb_const), finv(comb_const), inv(comb_const);
bool COMineted = false;
void COMinit() {

  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (ll i = 2; i < comb_const; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }

  COMineted = true;
}

// 二項係数計算
ll COM(ll n, ll k) {
  if (COMineted == false)
    COMinit();

  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//---------------------------------------------------------------
// 繰り返し２乗法 base^sisuu
//---------------------------------------------------------------
ll RepeatSquaring(ll base, ll sisuu) {
  if (sisuu < 0) {
    cout << "RepeatSquaring: 指数が負です!" << endl;
    return 0;
  }
  if (sisuu == 0)
    return 1;
  if (sisuu % 2 == 0) {
    ll t = RepeatSquaring(base, sisuu / 2) % MOD;
    return (t * t) % MOD;
  }
  return base * RepeatSquaring(base, sisuu - 1) % MOD;
}

//---------------------------------------------------------------
// 高速単発コンビネーション計算(O(logN))
//---------------------------------------------------------------
ll fast_com(ll a, ll b) {

  ll bunshi = 1;
  ll bunbo = 1;
  for (ll i = 1; i <= b; i++) {
    bunbo *= i;
    bunbo %= MOD;

    bunshi *= (a - i + 1);
    bunshi %= MOD;
  }

  ll ret = bunshi * RepeatSquaring(bunbo, MOD - 2);

  ret %= MOD;
  while (ret < 0) {
    ret += MOD;
  }

  return ret;
}

//---------------------------------------------------------------
// 2直線の交差判定(直線(x1, y1)->(x2, y2) と 直線(X1, Y1)->(X2, Y2))
//---------------------------------------------------------------
bool is_cross(ll x1, ll y1, ll x2, ll y2, ll X1, ll Y1, ll X2, ll Y2) {
  ll dx_ai = X1 - x1;
  ll dy_ai = Y1 - y1;
  ll dx_bi = X1 - x2;
  ll dy_bi = Y1 - y2;

  ll dx_ai2 = X2 - x1;
  ll dy_ai2 = Y2 - y1;
  ll dx_bi2 = X2 - x2;
  ll dy_bi2 = Y2 - y2;

  ll si = dx_ai * dy_bi - dy_ai * dx_bi;
  ll si2 = dx_ai2 * dy_bi2 - dy_ai2 * dx_bi2;

  ll si3 = dx_ai * dy_ai2 - dy_ai * dx_ai2;
  ll si4 = dx_bi * dy_bi2 - dy_bi * dx_bi2;

  return (si * si2 < 0 && si3 * si4 < 0);
}

//---------------------------------------------------------------
// 最長増加部分列の長さ(O(NlogN))
//---------------------------------------------------------------
ll LSI(vll vec, ll size) {
  vll lsi(size + 1); // 長さjを作った時の右端の最小値
  for (ll i = 0; i <= size; i++) {
    lsi[i] = LLINF;
  }

  lsi[0] = 0;
  lsi[1] = vec[0];

  for (ll i = 1; i < size; i++) {
    // 初めてvec[i]の方が小さくなるところを探す
    auto Iter = lower_bound(lsi.begin(), lsi.end(), vec[i]);
    ll idx = Iter - lsi.begin();

    if (idx > 0 && lsi[idx - 1] < vec[i]) // 条件文の前半怪しい
    {
      lsi[idx] = vec[i];
    }
  }

  for (ll i = size; i >= 0; i--) {
    if (lsi[i] < LLINF) {
      return i;
    }
  }
}

//---------------------------------------------------------------
// 木の根からの深さ
//---------------------------------------------------------------
vll tree_depth(vvll edge, ll start_node, ll n_node) {
  vll dist(n_node, LLINF);
  dist[start_node] = 0;

  stack<pll> S;
  S.push({start_node, 0});

  while (!S.empty()) {
    ll node = S.top().first;
    ll d = S.top().second;
    dist[node] = d;

    S.pop();

    for (int i = 0; i < edge[node].size(); i++) {
      if (dist[edge[node][i]] == LLINF) {
        S.push({edge[node][i], d + 1});
      }
    }
  }

  return dist;
}

//---------------------------------------------------------------
// ワーシャルフロイド法(O(N^3)) 任意の2点間の最短距離
//---------------------------------------------------------------
vvll warshall_floyd(ll n, vvll d) {
  for (int k = 0; k < n; k++) {     // 経由する頂点
    for (int i = 0; i < n; i++) {   // 始点
      for (int j = 0; j < n; j++) { // 終点
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
  return d;
}

//---------------------------------------------------------------
// Union Find
//---------------------------------------------------------------
/*
        UnionFind uf(要素の個数);
        for(int i = 0;i < 関係の個数; i++)
        {
                uf.merge(A[i], B[i]);
        }

        nを含む集合の大きさ = uf.size(n)
        nを含む集合の代表者 = uf.root(n)
        集合の個数 = uf.n_group
*/
class UnionFind {
public:
  vector<ll> par; // 各元の親を表す配列
  vector<ll> siz; // 素集合のサイズを表す配列(1 で初期化)
  ll n_group;     // 集合の数

  // Constructor
  UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
    for (ll i = 0; i < sz_; ++i)
      par[i] = i; // 初期では親は自分自身
    n_group = sz_;
  }
  void init(ll sz_) {
    par.resize(sz_);
    siz.assign(sz_, 1LL); // resize だとなぜか初期化されなかった
    for (ll i = 0; i < sz_; ++i)
      par[i] = i; // 初期では親は自分自身
  }

  // Member Function
  // Find
  ll root(ll x) { // 根の検索
    while (par[x] != x) {
      x = par[x] = par[par[x]]; // x の親の親を x の親とする
    }
    return x;
  }

  // Union(Unite, Merge)
  bool merge(ll x, ll y) {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    // merge technique（データ構造をマージするテク．小を大にくっつける）
    if (siz[x] < siz[y])
      swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    n_group--;
    return true;
  }

  bool issame(ll x, ll y) { // 連結判定
    return root(x) == root(y);
  }

  ll size(ll x) { // 素集合のサイズ
    return siz[root(x)];
  }
};

//========================================================================

struct edge {
  int to, cost;
};
typedef vector<vector<edge>> AdjList;
AdjList graph;

typedef pair<int, int> P;

vector<int> dist;
vector<int> prever;

void dijkstra(int n, int s) {
  dist = vector<int>(n, INF);
  prever = vector<int>(n, -1);
  dist[s] = 0;

  priority_queue<P, vector<P>, greater<P>> que;
  que.push(P(0, s));

  while (!que.empty()) {
    P p = que.top();
    que.pop();
    int v = p.second;
    if (dist[v] < p.first) {
      continue;
    }
    for (int i = 0; i < graph[v].size(); i++) {
      edge e = graph[v][i];
      if (dist[e.to] > dist[v] + e.cost) {
        dist[e.to] = dist[v] + e.cost;
        prever[e.to] = v;
        que.push(P(dist[e.to], e.to));
      }
    }
  }
}

vector<int> get_path(int t) { // 頂点tへの最短路
  vector<int> path;
  for (; t != -1; t = prever[t]) {
    path.push_back(t);
  }

  reverse(path.begin(), path.end());
  return path;
}

int main(void) {
  int n, m;
  cin >> n >> m;
  graph = AdjList(n);
  for (int i = 0; i < m; i++) {
    int from, to, cost;
    cin >> from >> to;
    from--;
    to--;
    edge t = {to, 1};
    edge t2 = {from, 1};
    graph[from].push_back(t);
    graph[to].push_back(t2);
  }

  dijkstra(n, 0);

  vll ans(n, -1);
  ans[0] = -1;

  vbool ok(n, false);
  ok[0] = true;

  for (int i = 1; i < n; i++) {
    if (!ok[i]) {
      vint rout = get_path(i);
      reverse(rout.begin(), rout.end());

      for (int j = 0; j < rout.size() - 1; j++) {
        if (ans[rout[j]] == -1) {
          ans[rout[j]] = rout[j + 1];
        } else {
          break;
        }
      }
    }
  }

  for (int i = 1; i < n; i++) {
    if (ans[i] == -1) {
      cout << "No";
      return 0;
    }
  }

  cout << "Yes" << endl;
  for (int i = 1; i < n; i++) {
    cout << ans[i] + 1 << endl;
  }
}