結果

問題 No.2910 単体ホモロジー入門
ユーザー 👑 binap
提出日時 2024-10-04 23:28:59
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 21 ms / 2,000 ms
コード長 3,650 bytes
コンパイル時間 4,249 ms
コンパイル使用メモリ 254,684 KB
最終ジャッジ日時 2025-02-24 15:40:18
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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ファイルパターン 結果
other AC * 47
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>
#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;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
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; 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;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 
    1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
int main(){
    int n, m;
    cin >> n >> m;
    vector<vector<int>> G(n, vector<int>(n));
    rep(i, m){
        int u, v;
        cin >> u >> v;
        G[u][v] = 1;
        G[v][u] = 1;
    }
    vector<int> v(3);
    cin >> v;
    
    sort(v.begin(), v.end());
    
    string ans = "No\n";
    
    if(n >= 3){
        vector<int> perm(3);
        rep(i, 3) perm[i] = i;
        do{
            vector<int> perm_sorted = perm;
            sort(perm_sorted.begin(), perm_sorted.end());
            if(perm_sorted[0] == v[0] and perm_sorted[1] == v[1] and perm_sorted[2] == v[2]) continue;
            if(G[perm[0]][perm[1]] == 1 and G[perm[1]][perm[2]] == 1 and G[perm[2]][perm[0]] == 1) ans = "Yes\n";
        }while(next_permutation(perm.begin(), perm.end()));
    }
    if(n >= 4){
        vector<int> perm(4);
        rep(i, 4) perm[i] = i;
        do{
            vector<int> perm_sorted(3);
            rep(i, 3) perm_sorted[i] = perm[i];
            sort(perm_sorted.begin(), perm_sorted.end());
            if(perm_sorted[0] == v[0] and perm_sorted[1] == v[1] and perm_sorted[2] == v[2]) continue;
            if(G[perm[0]][perm[1]] == 1 and G[perm[1]][perm[2]] == 1 and G[perm[2]][perm[0]] == 1) ans = "Yes\n";
        }while(next_permutation(perm.begin(), perm.end()));
    }
    if(n >= 4){
        vector<int> perm(4);
        rep(i, 4) perm[i] = i;
        do{
            if(G[perm[0]][perm[1]] == 1 and G[perm[1]][perm[2]] == 1 and G[perm[2]][perm[3]] == 1 and G[perm[3]][perm[0]] == 1) ans = "Yes\n";
        }while(next_permutation(perm.begin(), perm.end()));
    }
    cout << ans;
    
    return 0;
}
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