結果

問題 No.2832 Nana's Fickle Adventure
コンテスト
ユーザー 👑 binap
提出日時 2024-08-03 01:43:51
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 117 ms / 3,000 ms
コード長 4,790 bytes
コンパイル時間 4,126 ms
コンパイル使用メモリ 259,368 KB
最終ジャッジ日時 2025-02-23 20:42:48
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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ファイルパターン 結果
other AC * 48
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ソースコード

diff #
raw source code 

#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> 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);}

using mint = modint998244353;

// https://youtu.be/ylWYSurx10A?t=2352
template<typename T>
struct Matrix  : vector<vector<T>> {
	int h, w;
	Matrix(int h, int w, T val=0): vector<vector<T>>(h, vector<T>(w, val)), h(h), w(w) {}
	Matrix(initializer_list<initializer_list<T>> a) : vector<vector<T>>(a.begin(), a.end()){
		assert(int(this->size()) >= 1);
		assert(int((*this)[0].size()) >= 1);
		h = this->size();
		w = (*this)[0].size();
		rep(i, h) assert(int((*this)[i].size()) == w);
	}
	Matrix& unit() {
		assert(h == w);
		rep(i,h) (*this)[i][i] = 1;
		return *this;
	}
	Matrix operator*=(const Matrix& M){
		assert(w == M.h);
		Matrix r(h, M.w);
		rep(i,h) rep(k,w) rep(j, M.w){
			r[i][j] += (*this)[i][k] * M[k][j];
		}
		swap(*this, r);
		return *this;
	}
	Matrix operator*(const Matrix& M) const {return (Matrix(*this) *= M);}
	Matrix operator*=(const T& a) {
		for(int i = 0; i < h; i++) for(int j = 0; j < w; j++) (*this)[i][j] *= a;
		return *this;
	}
	Matrix operator*(const T& a) const {return (Matrix(*this) *= a);}
	Matrix operator+=(const Matrix& M){
		assert(h == M.h and w == M.w);
		for(int i = 0; i < h; i++) for(int j = 0; j < w; j++) (*this)[i][j] += M[i][j];
		return *this;
	}
	Matrix operator+(const Matrix& M) const {return (Matrix(*this) += M);}
	Matrix pow(long long t) const {
		assert(h == w);
		if (!t) return Matrix(h,h).unit();
		if (t == 1) return *this;
		Matrix r = pow(t>>1);r = r*r;
		if (t&1) r = r*(*this);
		return r;
	}
};

int main(){
	int n, m, t;
	cin >> n >> m >> t;
	vector<vector<int>> path(n, vector<int>(n));
	vector<int> path_tot(n);
	auto f = [&](int u, int v){
		return u * n + v;
	};
	rep(i, m){
		int u, v;
		cin >> u >> v;
		u--; v--;
		if(u != v){
			path[u][v]++;
			path[v][u]++;
			path_tot[u]++;
			path_tot[v]++;
		}else{
			path[u][u]++;
			path_tot[u]++;
		}
	}
	int k = (n + 1) * n;
	Matrix<mint> mat(k, k);
	rep(prev, n + 1) rep(from, n){
		if(path_tot[from] == 0){
			mat[f(n, from)][f(prev, from)] = 1;
			continue;
		}
		if(path_tot[from] == 1 and prev != n){
			mat[f(n, from)][f(prev, from)] = 1;
			continue;
		}
		rep(to, n){
			mint prob = 1;
			if(path[from][to] == 0) continue;
			if(prev == n){
				prob *= path[from][to];
				prob /= path_tot[from];
			}else if(prev != n and prev == to){
				prob *= path[from][to] - 1;
				prob /= path_tot[from] - 1;
			}else if(prev != n and prev != to){
				prob *= path[from][to];
				prob /= path_tot[from] - 1;
			}
			mat[f(from, to)][f(prev, from)] += prob;
		}
	}
	auto mat_pow = mat.pow(t);
	vector<mint> ans(n);
	rep(from, n + 1) rep(to, n){
		ans[to] += mat_pow[f(from, to)][f(n, 0)];
	}
	rep(to, n){
		cout << ans[to] << "\n";
	}
	return 0;
}
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