結果

問題 No.2674 k-Walk on Bipartite
ユーザー Daylight
提出日時 2024-03-15 23:17:18
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 203 ms / 2,000 ms
コード長 3,862 bytes
コンパイル時間 2,442 ms
コンパイル使用メモリ 214,908 KB
最終ジャッジ日時 2025-02-20 06:04:10
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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 * 3
other AC * 36
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 <atcoder/dsu>
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) (int) (x).size()
#define REP(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(auto i = (a); i < (b); i++)
#define For(i, a, b, c) \
    for(auto i = (a); i != (b); i += (c))
#define REPR(i, n) for(auto i = (n) -1; i >= 0; i--)
#define ALL(s) (s).begin(), (s).end()
#define so(V) sort(ALL(V))
#define rev(V) reverse(ALL(V))
#define uni(v) v.erase(unique(ALL(v)), (v).end())
#define eb emplace_back
typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> PI;
typedef pair<ll, ll> PL;
const double EPS = 1e-6;
const int MOD = 1e9 + 7;
const int INF = (1 << 30);
const ll LINF = 1e18;
const double math_PI = acos(-1);
template<typename T>
vector<T> make_v(size_t a) {
    return vector<T>(a);
}
template<typename T, typename... Ts>
auto make_v(size_t a, Ts... ts) {
    return vector<decltype(make_v<T>(ts...))>(
        a, make_v<T>(ts...));
}
template<typename T, typename V>
typename enable_if<is_class<T>::value == 0>::type fill_v(
    T& t, const V& v) {
    t = v;
}
template<typename T, typename V>
typename enable_if<is_class<T>::value != 0>::type fill_v(
    T& t, const V& v) {
    for(auto& e: t) fill_v(e, v);
}
template<class T>
bool chmax(T& a, const T& b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
bool chmin(T& a, const T& b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template<typename S, typename T>
istream& operator>>(istream& is, pair<S, T>& p) {
    cin >> p.first >> p.second;
    return is;
}
template<typename T>
istream& operator>>(istream& is, vector<T>& vec) {
    for(T& x: vec) is >> x;
    return is;
}
template<typename T>
string join(vector<T>& vec, string splitter) {
    stringstream ss;
    REP(i, SZ(vec)) {
        if(i != 0) ss << splitter;
        ss << vec[i];
    }
    return ss.str();
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
    os << join(vec, " ");
    return os;
}
using namespace atcoder;
template<class T = ll>
struct Edge {
public:
    int from, to;
    T cost;
    Edge() {
    }
    Edge(int _from, int _to, T _cost) {
        from = _from;
        to = _to;
        cost = _cost;
    }
};
template<class T = ll>
using Edges = vector<Edge<T>>;
template<class T = ll>
using Graph = vector<Edges<T>>;
vll dijkstra(int s, const Graph<ll>& G) {
    using P = pair<ll, int>;
    priority_queue<P, vector<P>, greater<>> que;
    vll D(SZ(G), LINF);
    D[s] = 0;
    que.push(P(0, s));
    while(!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second;
        if(D[v] < p.first) continue;
        for(Edge e: G[v]) {
            if(D[e.to] > D[v] + e.cost) {
                D[e.to] = D[v] + e.cost;
                que.push(P(D[e.to], e.to));
            }
        }
    }
    return D;
}
int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    int s, t, k;
    cin >> s >> t >> k;
    s--;
    t--;
    Graph<> G(N);
    dsu D(N);
    REP(i, M) {
        int A, B;
        cin >> A >> B;
        A--;
        B--;
        G[A].eb(A, B, 1);
        G[B].eb(B, A, 1);
        D.merge(A, B);
    }
    int g = SZ(D.groups());
    if(N == 1) {
        cout << "No" << endl;
        return 0;
    }
    if(s == t) {
        if(k % 2 == 1) {
            if(g >= 2 && N >= 3) {
                cout << "Unknown" << endl;
            } else {
                cout << "No" << endl;
            }
            return 0;
        } else {
            if(SZ(G[s]) == 0) {
                cout << "Unknown" << endl;
            } else {
                cout << "Yes" << endl;
            }
        }
        return 0;
    }
    auto d = dijkstra(s, G);
    if(!D.same(s, t)) {
        if(D.size(s) == 1 && D.size(t) == 1) {
            if(k % 2 == 1) {
                cout << "Unknown" << endl;
            } else if(g >= 3) {
                cout << "Unknown" << endl;
            } else {
                cout << "No" << endl;
            }
        } else {
            cout << "Unknown" << endl;
        }
    } else if(d[t] % 2 != k % 2) {
        cout << "No" << endl;
    } else if(d[t] > k) {
        cout << "Unknown" << endl;
    } else {
        cout << "Yes" << endl;
    }
    return 0;
}
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