矩阵的k次方在图论中的意义

题目：武汉工程大学第八届ACM新生赛 D

一个矩阵的k次方后，a[i][j]表示从i到j长度为k的条数

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using i128=__int128;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using db = double;
using ldb = long double;
#define F first
#define S second
#define int ll
const int MOD=998244353;
ll ksm(ll a,ll x,ll mod=MOD){
    a%=mod;
    if(a==0) return 0;
    ll ans=1;
    while(x){
        if(x&1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        x>>=1;
    }
    return ans;
}
struct matrix{
    vector<vector<ll>> a;
    int n;
    matrix(int x,bool _=false){
        n=x;
        a.assign(n+1,vector<ll>(n+1,0));
        if(_){
            for(int i=1;i<=n;i++) a[i][i]=1;
        }
    }
    matrix operator*(const matrix &b)const{
        matrix res(n);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                for(int k=1;k<=n;k++){
                    res.a[i][j]=(res.a[i][j]+a[i][k]*b.a[k][j]%(MOD))%(MOD);
                }
            }
        }
        return res;
    }
    static matrix ksm(matrix x,ll y){
        matrix res(x.n,true);
        while(y){
            if(y&1) res=res*x;
            x=x*x;
            y>>=1;
        }
        return res;
    }
};
void solve(){
    ll n;cin>>n;
    if(n==1){
        cout<<"5\n";
        return;
    }
    matrix base(5);
    for(int i=1;i<=5;i++){
        int k;cin>>k;
        for(int j=1;j<=k;j++){
            int x;cin>>x;
            base.a[i][x]=1;
        }
    }
    matrix ans=matrix::ksm(base,n-1);
    ll res=0;
    for(int i=1;i<=5;i++){
        for(int j=1;j<=5;j++){
            res=(res+ans.a[i][j])%MOD;
        }
    }
    cout<<res<<"\n";
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr),cout.tie(nullptr);
    int t=1;
    // cin>>t;
    while(t--) solve();
    return 0;
}