結果
問題 | No.2536 同値性と充足可能性 |
ユーザー |
👑 |
提出日時 | 2023-11-10 21:56:09 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 149 ms / 2,000 ms |
コード長 | 2,241 bytes |
コンパイル時間 | 4,751 ms |
コンパイル使用メモリ | 262,320 KB |
最終ジャッジ日時 | 2025-02-17 20:43:10 |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 31 |
ソースコード
#include<bits/stdc++.h>#include<atcoder/all>#define rep(i,n) for(int i=0;i<n;i++)using namespace std;using namespace atcoder;typedef long long ll;typedef vector<int> vi;typedef vector<long long> vl;typedef vector<vector<int>> vvi;typedef vector<vector<long long>> vvl;typedef long double ld;typedef pair<int, int> P;ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second << "\n"; return os;}template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); returnos;}template <typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : "");return os;}int main(){int n, m;cin >> n >> m;dsu uf(n);vector<vector<pair<int, int>>> G(n);rep(i, m){int u, v;string s;cin >> u >> s >> v;u--; v--;int e = -1;if(s == "<=/=>") e = 1;if(s == "<==>") e = 0;G[u].emplace_back(v, e);G[v].emplace_back(u, e);uf.merge(u, v);}vi c(n, -1);bool success = true;auto dfs = [&](int from, int prev, auto dfs) -> void{for(auto [to, e] : G[from]){if(to == prev) continue;if(c[to] != -1){if(c[to] != c[from] ^ e) success = false;continue;}c[to] = c[from] ^ e;dfs(to, from, dfs);}};rep(i, n){if(c[i] == -1){c[i] = 0;dfs(i, -1, dfs);}}if(!success){cout << "No\n";return 0;}auto groups = uf.groups();int mx = 0;vi ans;for(auto& v : groups){vi d(2);for(int x : v) d[c[x]]++;if(d[0] >= d[1]){mx += d[0];for(int x : v) if(c[x] == 0) ans.push_back(x + 1);}else{mx += d[1];for(int x : v) if(c[x] == 1) ans.push_back(x + 1);}}sort(ans.begin(), ans.end());if(mx * 2 >= n){cout << "Yes\n";cout << mx << "\n";cout << ans;}else cout << "No\n";return 0;}