結果

問題 No.2674 k-Walk on Bipartite
ユーザー 👑 binap
提出日時 2024-03-15 22:12:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 388 ms / 2,000 ms
コード長 3,768 bytes
コンパイル時間 4,500 ms
コンパイル使用メモリ 267,528 KB
最終ジャッジ日時 2025-02-20 05:19:06
ジャッジサーバーID
（参考情報） 		judge1 / 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 * 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 # 

#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);
	}
	if(n == 1){
		cout << "No\n";
		return 0;
	}
	if(s == t){
		if(k % 2) cout << "No\n";
		else{
			if(G[s].size() > 0) cout << "Yes\n";
			else cout << "Unknown\n";
		}
		return 0;
	}
	if(n == 2){
		assert(s != t);
		if(k % 2){
			if(m == 1) cout << "Yes\n";
			else cout << "Unknown\n";
		}else cout << "No\n";
		return 0;
	}
	int dist = graph.get_dist(s, t);
	if(dist >= INF){
		cout << "Unknown\n";
	}else{
		int dif = abs(dist - k);
		if(dif % 2) cout << "No\n";
		else{
			if(dist <= k) cout << "Yes\n";
			else cout << "Unknown\n";
		}
	}
	return 0;
}
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