結果

問題 No.2674 k-Walk on Bipartite
ユーザー 👑 binap
提出日時 2024-03-15 22:00:02
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,543 bytes
コンパイル時間 4,685 ms
コンパイル使用メモリ 268,680 KB
最終ジャッジ日時 2025-02-20 05:06:52
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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 * 32 WA * 4
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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 <int m> ostream& operator<<(ostream& os, const dynamic_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; 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> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
template<typename T>
struct Edge_Dijkstra{
    int from, to;
    T cost;
    Edge_Dijkstra(int from, int to, T cost) : from(from), to(to), cost(cost) {};
};
const int INF = 1001001001;
template<typename T>
struct Dijkstra{
    int n, m;
    vector<bool> initialized;
    vector<Edge_Dijkstra<T>> E;
    vector<vector<int>> G;
    map<int, vector<T>> dist;
    map<int, vector<int>> idx;
    Dijkstra(int _n) : n(_n), m(0), initialized(n, false), G(n){}
    void add_edge(int from, int to, T cost){
        Edge_Dijkstra e(from, to, cost);
        E.push_back(e);
        G[from].emplace_back(m);
        m++;
    }
    void calc(int s){
        initialized[s] = true;
        dist[s] = vector<T>(n, INF);
        idx[s] = vector<int>(n, -1);
        priority_queue<tuple<T, int, int>, vector<tuple<T, int, int>>, greater<tuple<T, int, int>>> pq;
        pq.emplace(0, s, -1);
        while(pq.size()){
            auto [cost, from, index] = pq.top(); pq.pop();
            if(dist[s][from] <= cost) continue;
            dist[s][from] = cost;
            idx[s][from] = index;
            for(int index : G[from]){
                int to = E[index].to;
                T cost_plus = E[index].cost;
                if(dist[s][to] <= cost + cost_plus) continue;
                pq.emplace(cost + cost_plus, to, index);
            }
        }
    }
    int farthest(int s){
        if(!initialized[s]) calc(s);
        int idx = 0;
        rep(i, n) if(dist[s][i] > dist[s][idx]) idx = i;
        return idx;
    }
    T get_dist(int s, int t){
        if(!initialized[s]) calc(s);
        return dist[s][t];
    }
    vi restore(int s, int t){
        if(!initialized[s]) calc(s);
        if(dist[s][t] == INF) return vi(0);
        vi res;
        while(idx[s][t] != -1){
            auto e = E[idx[s][t]];
            res.push_back(idx[s][t]);
            t = e.from;
        }
    reverse(res.begin(), res.end());
    return res;
    }
};
int main(){
    int n, m;
    cin >> n >> m;
    int s, t, k;
    cin >> s >> t >> k;
    s--; t--;
    vvi G(n);
    Dijkstra<int> graph(n);
    rep(i, m){
        int u, v;
        cin >> u >> v;
        u--; v--;
        graph.add_edge(u, v, 1);
        graph.add_edge(v, u, 1);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    int dist = graph.get_dist(s, t);
    if(dist >= INF){
        cout << "Unknown\n";
    }else{
        int diff = abs(dist - k);
        if(diff % 2) cout << "No\n";
        else{
            if(s == t){
                if(G[s].size() >= 1) cout << "Yes\n";
                else cout << "No\n";
            }else{
                if(dist <= k) cout << "Yes\n";
                else cout << "Unknown\n";
            }
        }
    }
    return 0;
}
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