結果

問題 No.3105 Parallel Connection and Spanning Trees
ユーザー umimel
提出日時 2025-04-11 23:24:02
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 545 ms / 5,000 ms
コード長 7,370 bytes
コンパイル時間 1,996 ms
コンパイル使用メモリ 175,008 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-04-11 23:24:11
合計ジャッジ時間 9,517 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
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 * 2
other AC * 32
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>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
#define all(a) (a).begin(), (a).end()
#define pb push_back
#define fi first
#define se second
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
const ll MOD1000000007 = 1000000007;
const ll MOD998244353 = 998244353;
const ll MOD[3] = {999727999, 1070777777, 1000000007};
const ll LINF = 1LL << 60LL;
const int IINF = (1 << 30) - 1;


template<long long mod>
class modint{
    long long x;
public:
    modint(long long x=0) : x((x%mod+mod)%mod) {}
    modint operator-() const { 
      return modint(-x);
    }
    bool operator==(const modint& a){
        if(x == a) return true;
        else return false;
    }
    bool operator==(long long a){
        if(x == a) return true;
        else return false;
    }
    bool operator!=(const modint& a){
        if(x != a) return true;
        else return false;
    }
    bool operator!=(long long a){
        if(x != a) return true;
        else return false;
    }
    modint& operator+=(const modint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    modint& operator-=(const modint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    modint& operator*=(const  modint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    modint operator+(const modint& a) const {
        modint res(*this);
        return res+=a;
    }
    modint operator-(const modint& a) const {
        modint res(*this);
        return res-=a;
    }
    modint operator*(const modint& a) const {
        modint res(*this);
        return res*=a;
    }
    modint pow(long long t) const {
        if (!t) return 1;
        modint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime mod
    modint inv() const {
        return pow(mod-2);
    }
    modint& operator/=(const modint& a) {
        return (*this) *= a.inv();
    }
    modint operator/(const modint& a) const {
        modint res(*this);
        return res/=a;
    }

    friend std::istream& operator>>(std::istream& is, modint& m) noexcept {
        is >> m.x;
        m.x %= mod;
        if (m.x < 0) m.x += mod;
        return is;
    }

    friend ostream& operator<<(ostream& os, const modint& m){
        os << m.x;
        return os;
    }
};


template<typename T>
struct matrix{
    vector<vector<T>> A;

    matrix(){}
    matrix(size_t n, size_t m) : A(n, vector<T>(m, 0)){}
    matrix(size_t n) : A(n, vector<T>(n, 0)){};

    size_t height() const{return (A.size());}
    size_t width() const{return (A[0].size());}

    inline const vector<T> &operator[](int k) const{return (A.at(k));}

    inline vector<T> &operator[](int k){return (A.at(k));}

    static matrix I(size_t n){
        matrix mat(n);
        for(int i=0; i<n; i++) mat[i][i] = 1;
        return (mat);
    }

    matrix &operator+=(const matrix &B){
        size_t n = height(), m = width();
        assert(n == B.height() && m == B.width());
        for(int i=0; i<n; i++)for(int j=0; j<m; j++) (*this)[i][j] += B[i][j];
        return (*this);
    }

    matrix &operator-=(const matrix &B){
        size_t n = height(), m = width();
        assert(n == B.height() && m == B.width());
        for(int i=0; i<n; i++)for(int j=0; j<m; j++) (*this)[i][j] -= B[i][j];
        return (*this);
    }

    matrix &operator*=(const matrix &B){
        size_t n = height(), m = B.width(), p = width();
        assert(p == B.height());
        vector<vector<T>> C(n, vector<T>(m, 0));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                for(int k = 0; k < p; k++)
                    C[i][j] = (C[i][j] + (*this)[i][k] * B[k][j]);
        A.swap(C);
        return (*this);
    }

    matrix &operator^=(long long k){
        matrix B = matrix::I(height());
        while(k > 0) {
            if(k & 1) B *= *this;
            *this *= *this;
            k >>= 1LL;
        }
        A.swap(B.A);
        return (*this);
    }

    matrix operator+(const matrix &B) const{
        return (matrix(*this) += B);
    }

    matrix operator-(const matrix &B) const{
        return (matrix(*this) -= B);
    }

    matrix operator*(const matrix &B) const{
        return (matrix(*this) *= B);
    }

    matrix operator^(const long long k) const{
        return (matrix(*this) ^= k);
    }

    friend ostream &operator<<(ostream &os, matrix &p){
        size_t n = p.height(), m = p.width();
        for(int i=0; i<n; i++){
            os << "[";
            for(int j=0; j<m; j++){
                os << p[i][j] << (j + 1 == m ? "]\n" : ",");
            }
        }
        return (os);
    }


    T determinant(){
        matrix B(*this);
        assert(width() == height());
        T ret = 1;
        for(int i=0; i<width(); i++) {
            int idx = -1;
            for(int j=i; j<width(); j++) {
                if(B[j][i] != 0) idx = j;
            }
            if(idx == -1) return (0);
            if(i != idx) {
                ret *= -1;
                swap(B[i], B[idx]);
            }
            ret *= B[i][i];
            T vv = B[i][i];
            for(int j=0; j<width(); j++) {
                B[i][j] /= vv;
            }
            for(int j=i+1; j<width(); j++) {
                T a = B[j][i];
                for(int k=0; k<width(); k++) {
                    B[j][k] -= B[i][k] * a;
                }
            }
        }
        return (ret);
    }
};

using mint = modint<MOD998244353>;

template<typename T>
T count_spanning_tree(vector<vector<int>> &G){
    int n = (int)G.size();
    if(n==1) return T(1);

    matrix<T> L(n);
    for(int v=0; v<n; v++){
        L[v][v] = T((int)G[v].size());
        for(int to : G[v]) L[v][to]-=T(1);
    }

    matrix<T> L11(n-1);
    for(int i=0; i<n-1; i++)for(int j=0; j<n-1; j++) L11[i][j]=L[i+1][j+1];
    return L11.determinant();
}

void solve(){
    int k; cin >> k;
    vector<mint> cnt1(k, 0), cnt2(k, 0);
    for(int i=0; i<k; i++){
        int n, m; cin >> n >> m;
        vector<vector<int>> G1(n), G2(n-1);
        for(int i=0; i<m; i++){
            int u, v; cin >> u >> v;
            u--; v--;
            G1[u].push_back(v);
            G1[v].push_back(u);
        }
        for(int v=0; v<n; v++){
            for(int to : G1[v]){
                if(v==0||v==1){
                    if(to==0||to==1) continue;
                    G2[0].pb(to-1);
                }else{
                    if(to==0||to==1){
                        G2[v-1].pb(0);
                    }else{
                        G2[v-1].pb(to-1);
                    }
                }
            }
        }
        cnt1[i] = count_spanning_tree<mint>(G1);
        cnt2[i] = count_spanning_tree<mint>(G2);
    }

    vector<vector<int>> G(2*k+2);
    int x = 2*k, y = 2*k+1;
    for(int i=0; i<k; i++){
        G[x].pb(2*i);
        G[2*i].pb(x);
        G[y].pb(2*i+1);
        G[2*i+1].pb(y);
    }

    mint ans = 0;
    for(int i=0; i<k; i++){
        mint sa = cnt1[i];
        for(int j=0; j<k; j++){
            if(i==j) continue;
            sa *= cnt2[j] + cnt1[j]*mint(2);
        }

        ans += sa;
    }

    cout << ans << '\n';
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    
    int T=1;
    //cin >> T;
    while(T--) solve();
}
0