結果

問題 No.274 The Wall
ユーザー Haar
提出日時 2019-04-12 12:30:31
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 931 ms / 2,000 ms
コード長 4,283 bytes
コンパイル時間 2,087 ms
コンパイル使用メモリ 184,176 KB
実行使用メモリ 388,992 KB
最終ジャッジ日時 2024-06-22 02:26:32
合計ジャッジ時間 5,216 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 22
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#define FOR(v, a, b) for(int v = (a); v < (b); ++v)
#define FORE(v, a, b) for(int v = (a); v <= (b); ++v)
#define REP(v, n) FOR(v, 0, n)
#define REPE(v, n) FORE(v, 0, n)
#define REV(v, a, b) for(int v = (a); v >= (b); --v)
#define ALL(x) (x).begin(), (x).end()
#define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it)
#define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it)
#define EXIST(c,x) ((c).find(x) != (c).end())
#define LLI long long int
#define fst first
#define snd second
#ifdef DEBUG
#include <misc/C++/Debug.cpp>
#else
#define dump(x) ((void)0)
#endif
#define gcd __gcd
using namespace std;
template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;}
template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;}
template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;}
template <typename T, typename U> istream& operator>>(istream &is, pair<T,U> &p){is >> p.first >> p.second; return is;}
template <typename T, typename U> bool chmin(T &a, const U &b){return (a>b ? a=b, true : false);}
template <typename T, typename U> bool chmax(T &a, const U &b){return (a<b ? a=b, true : false);}
template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);}
template <typename Cost = int> class Edge{
public:
  int from,to;
  Cost cost;
  Edge() {}
  Edge(int from, int to, Cost cost): from(from), to(to), cost(cost){}
  Edge rev() const {return Edge(to,from,cost);}
  
  static bool cmp_to_lt(const Edge &e1, const Edge &e2){return e1.to < e2.to;}
  static bool cmp_cost_lt(const Edge &e1, const Edge &e2){return e1.cost < e2.cost;}
  static bool cmp_to_gt(const Edge &e1, const Edge &e2){return e1.to > e2.to;}
  static bool cmp_cost_gt(const Edge &e1, const Edge &e2){return e1.cost > e2.cost;}
  friend ostream& operator<<(ostream &os, const Edge &e){
    os << "(FROM: " << e.from << "," << "TO: " << e.to << "," << "COST: " << e.cost << ")";
    return os;
  }
};
template <typename T> using Graph = vector<vector<Edge<T>>>;
template <typename T> vector<int> strongly_connected_components(Graph<T> &graph){
  int n = graph.size();
  vector<bool> visit(n);
  vector<int> check(n);
  function<void(int)> dfs =
    [&](int cur){
      visit[cur] = true;
      for(auto &e : graph[cur]) if(!visit[e.to]) dfs(e.to);
      check.push_back(cur);
    };
  
  REP(i,n) if(!visit[i]) dfs(i);
  Graph<T> rgraph(n);
  REP(i,n) for(auto &e : graph[i]) rgraph[e.to].push_back(e.rev());
  vector<int> ret(n,-1);
  reverse(ALL(check));
  function<void(int,int)> rdfs =
    [&](int cur, int i){
      ret[cur] = i;
      for(auto &e : rgraph[cur]) if(ret[e.to] == -1) rdfs(e.to,i);
    };
  int i = 0;
  for(auto c : check) if(ret[c] == -1) {rdfs(c,i); ++i;}
    
  return ret;
}
class two_sat{
  int n;
  Graph<int> g;
  
public:
  two_sat(int n): n(n), g(2*n){}
  int inv(int i){ // not
    if(i<n) return i+n;
    else return i-n;
  }
  void add(int a, int b){
    if(a == b){ // a ∨ a <=> (a => !a)
      g[a].push_back(Edge<int>(a, inv(a), 1));
    }else{ // a ∨ b <=> (a => !b) ∧ (b => !a)
      g[a].push_back(Edge<int>(a, inv(b), 1));
      g[b].push_back(Edge<int>(b, inv(a), 1));
    }
  }
  bool solve(){
    auto s = strongly_connected_components(g);
    REP(i,n){
      if(s[i] == s[i+n]) return false;
    }
    return true;
  }
};
bool intersect(int l1, int r1, int l2, int r2){
  if(r1 < l2 or r2 < l1) return false;
  else return true;
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N,M;
  while(cin >> N >> M){
    vector<int> L(N), R(N);
    REP(i,N) cin >> L[i] >> R[i];
    vector<int> L2(N), R2(N);
    REP(i,N){
      R2[i] = M - L[i] - 1;
      L2[i] = M - R[i] - 1;
    }
    two_sat sat(N);
    
    REP(i,N){
      FOR(j,i+1,N){
    if(intersect(L[i],R[i],L[j],R[j])) sat.add(sat.inv(i), sat.inv(j));
    if(intersect(L[i],R[i],L2[j],R2[j])) sat.add(sat.inv(i), j);
    if(intersect(L2[i],R2[i],L[j],R[j])) sat.add(i, sat.inv(j));
    if(intersect(L2[i],R2[i],L2[j],R2[j])) sat.add(i,j);
      }
    }
    bool ans = sat.solve();
    cout << (ans ? "YES" : "NO") << endl;
  }
  
  return 0;
}
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