結果

問題 No.2536 同値性と充足可能性
ユーザー 👑 binap
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
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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>
#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" : " "); return
    os;}
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;
}
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