結果
問題 | No.274 The Wall |
ユーザー |
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提出日時 | 2020-09-17 01:07:53 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 493 ms / 2,000 ms |
コード長 | 7,599 bytes |
コンパイル時間 | 2,813 ms |
コンパイル使用メモリ | 212,364 KB |
最終ジャッジ日時 | 2025-01-14 15:31:26 |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 22 |
ソースコード
#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef pair<ll, int> pli;typedef pair<int, int> pii;typedef pair<ll, ld> pld;typedef pair<pii, int> ppiii;typedef pair<pii, ll> ppiil;typedef pair<pll, ll> pplll;typedef pair<pli, int> pplii;typedef vector<vector<ll>> vvl;typedef vector<vector<int>> vvi;typedef vector<vector<pll>> vvpll;#define rep(i, a, b) for (ll i=(a); i<(b); i++)#define rrep(i, a, b) for (ll i=(a); i>(b); i--)#define pb push_back#define tostr to_string#define mkp make_pair#define list2d(name, N, M, type, init) vector<vector<type>> name(N, vector<type>(M, init))const ll INF = LONG_LONG_MAX;const ll MOD = 998244353;void print(ld out) { cout << fixed << setprecision(15) << out << '\n'; }void print(double out) { cout << fixed << setprecision(15) << out << '\n'; }template<typename T> void print(T out) { cout << out << '\n'; }template<typename T1, typename T2> void print(pair<T1, T2> out) { cout << out.first << ' ' << out.second << '\n'; }template<typename T> void print(vector<T> A) { rep(i, 0, A.size()) { cout << A[i]; cout << (i == A.size()-1 ? '\n' : ' '); } }template<typename T> void print(set<T> S) { vector<T> A(S.begin(), S.end()); print(A); }template<typename T> inline bool chmax(T &x, T y) { return (y > x) ? x = y, true : false; }template<typename T> inline bool chmin(T &x, T y) { return (y < x) ? x = y, true : false; }ll sum(vector<ll> A) { ll res = 0; for (ll a: A) res += a; return res; }ll max(vector<ll> A) { ll res = -INF; for (ll a: A) chmax(res, a); return res; }ll min(vector<ll> A) { ll res = INF; for (ll a: A) chmin(res, a); return res; }ll toint(string s) { ll res = 0; for (char c : s) { res *= 10; res += (c - '0'); } return res; }// 数字なら'0'、アルファベットなら'a'みたいに使い分ける// int toint(char c) { return c - '0'; }// char tochar(int i) { return '0' + i; }inline ll pow(int x, ll n) { ll res = 1; rep(_, 0, n) res *= x; return res; }inline ll pow(ll x, ll n, int mod) { ll res = 1; while (n > 0) { if (n & 1) { res = (res * x) % mod; } x = (x * x) % mod; n >>= 1; } return res; }inline ll floor(ll a, ll b) { if (a < 0) { return (a-b+1) / b; } else { return a / b; } }inline ll ceil(ll a, ll b) { if (a >= 0) { return (a+b-1) / b; } else { return a / b; } }pll divmod(ll a, ll b) { ll d = a / b; ll m = a % b; return {d, m}; }int popcount(ll S) { return __builtin_popcountll(S); }ll gcd(ll a, ll b) { return __gcd(a, b); }#ifndef ATCODER_INTERNAL_SCC_HPP#define ATCODER_INTERNAL_SCC_HPP 1#include <algorithm>#include <utility>#include <vector>namespace atcoder {namespace internal {template <class E> struct csr {std::vector<int> start;std::vector<E> elist;csr(int n, const std::vector<std::pair<int, E>>& edges): start(n + 1), elist(edges.size()) {for (auto e : edges) {start[e.first + 1]++;}for (int i = 1; i <= n; i++) {start[i] += start[i - 1];}auto counter = start;for (auto e : edges) {elist[counter[e.first]++] = e.second;}}};// Reference:// R. Tarjan,// Depth-First Search and Linear Graph Algorithmsstruct scc_graph {public:scc_graph(int n) : _n(n) {}int num_vertices() { return _n; }void add_edge(int from, int to) { edges.push_back({from, {to}}); }// @return pair of (# of scc, scc id)std::pair<int, std::vector<int>> scc_ids() {auto g = csr<edge>(_n, edges);int now_ord = 0, group_num = 0;std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);visited.reserve(_n);auto dfs = [&](auto self, int v) -> void {low[v] = ord[v] = now_ord++;visited.push_back(v);for (int i = g.start[v]; i < g.start[v + 1]; i++) {auto to = g.elist[i].to;if (ord[to] == -1) {self(self, to);low[v] = std::min(low[v], low[to]);} else {low[v] = std::min(low[v], ord[to]);}}if (low[v] == ord[v]) {while (true) {int u = visited.back();visited.pop_back();ord[u] = _n;ids[u] = group_num;if (u == v) break;}group_num++;}};for (int i = 0; i < _n; i++) {if (ord[i] == -1) dfs(dfs, i);}for (auto& x : ids) {x = group_num - 1 - x;}return {group_num, ids};}std::vector<std::vector<int>> scc() {auto ids = scc_ids();int group_num = ids.first;std::vector<int> counts(group_num);for (auto x : ids.second) counts[x]++;std::vector<std::vector<int>> groups(ids.first);for (int i = 0; i < group_num; i++) {groups[i].reserve(counts[i]);}for (int i = 0; i < _n; i++) {groups[ids.second[i]].push_back(i);}return groups;}private:int _n;struct edge {int to;};std::vector<std::pair<int, edge>> edges;};} // namespace internal} // namespace atcoder#endif // ATCODER_INTERNAL_SCC_HPP#ifndef ATCODER_TWOSAT_HPP#define ATCODER_TWOSAT_HPP 1#include <cassert>#include <vector>namespace atcoder {// Reference:// B. Aspvall, M. Plass, and R. Tarjan,// A Linear-Time Algorithm for Testing the Truth of Certain Quantified Boolean// Formulasstruct two_sat {public:two_sat() : _n(0), scc(0) {}two_sat(int n) : _n(n), _answer(n), scc(2 * n) {}void add_clause(int i, bool f, int j, bool g) {assert(0 <= i && i < _n);assert(0 <= j && j < _n);scc.add_edge(2 * i + (f ? 0 : 1), 2 * j + (g ? 1 : 0));scc.add_edge(2 * j + (g ? 0 : 1), 2 * i + (f ? 1 : 0));}bool satisfiable() {auto id = scc.scc_ids().second;for (int i = 0; i < _n; i++) {if (id[2 * i] == id[2 * i + 1]) return false;_answer[i] = id[2 * i] < id[2 * i + 1];}return true;}std::vector<bool> answer() { return _answer; }private:int _n;std::vector<bool> _answer;internal::scc_graph scc;};} // namespace atcoder#endif // ATCODER_TWOSAT_HPPusing namespace atcoder;ll N, M;vector<ll> L, R;bool check(ll l1, ll r1, ll l2, ll r2) {return (l1 >= r2 || l2 >= r1);}int main() {cin.tie(0);ios::sync_with_stdio(false);cin >> N >> M;L.resize(N);R.resize(N);rep(i, 0, N) {cin >> L[i] >> R[i];R[i]++;}// ts[i] := i行目を反転させるかtwo_sat ts(N);rep(i, 0, N) {rep(j, i+1, N) {// 0, 0if (!check(L[i], R[i], L[j], R[j])) {// a & bがNGなら、!a | !bを条件に追加するts.add_clause(i, 1, j, 1);}// 1, 0if (!check(M-R[i], M-L[i], L[j], R[j])) {ts.add_clause(i, 0, j, 1);}// 0, 1if (!check(L[i], R[i], M-R[j], M-L[j])) {ts.add_clause(i, 1, j, 0);}// 1, 1if (!check(M-R[i], M-L[i], M-R[j], M-L[j])) {ts.add_clause(i, 0, j, 0);}}}if (ts.satisfiable()) {print("YES");} else {print("NO");}return 0;}