結果

問題 No.3230 Mutual Corresponding System
ユーザー Taiki0715
提出日時 2025-08-08 22:31:04
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 310 ms / 2,000 ms
コード長 5,969 bytes
コンパイル時間 3,542 ms
コンパイル使用メモリ 296,332 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-08-08 22:31:18
合計ジャッジ時間 7,252 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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 * 23
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 ull=unsigned long long;
using P=pair<ll,ll>;
template<typename T>using minque=priority_queue<T,vector<T>,greater<T>>;
template<typename T>bool chmax(T &a,const T &b){return (a<b?(a=b,true):false);}
template<typename T>bool chmin(T &a,const T &b){return (a>b?(a=b,true):false);}
template<typename T1,typename T2>istream &operator>>(istream &is,pair<T1,T2>&p){is>>p.first>>p.second;return is;}
template<typename T1,typename T2,typename T3>istream &operator>>(istream &is,tuple<T1,T2,T3>&a){is>>std::get<0>(a)>>std::get<1>(a)>>std::get<2>(a);return is;}
template<typename T,size_t n>istream &operator>>(istream &is,array<T,n>&a){for(auto&i:a)is>>i;return is;}
template<typename T>istream &operator>>(istream &is,vector<T> &a){for(auto &i:a)is>>i;return is;}
template<typename T1,typename T2>void operator++(pair<T1,T2>&a,int n){a.first++,a.second++;}
template<typename T1,typename T2>void operator--(pair<T1,T2>&a,int n){a.first--,a.second--;}
template<typename T>void operator++(vector<T>&a,int n){for(auto &i:a)i++;}
template<typename T>void operator--(vector<T>&a,int n){for(auto &i:a)i--;}
#define overload3(_1,_2,_3,name,...) name
#define rep1(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define rep(...) overload3(__VA_ARGS__,rep2,rep1)(__VA_ARGS__)
#define reps(i,l,r) rep2(i,l,r)
#define all(x) x.begin(),x.end()
#define pcnt(x) __builtin_popcountll(x)
#define fin(x) return cout<<(x)<<'\n',static_cast<void>(0)
#define yn(x) cout<<((x)?"Yes\n":"No\n")
#define uniq(x) sort(all(x)),x.erase(unique(all(x)),x.end())
template<typename T>
inline int fkey(vector<T>&z,T key){return lower_bound(z.begin(),z.end(),key)-z.begin();}
ll myceil(ll a,ll b){return (a+b-1)/b;}
template<typename T,size_t n,size_t id=0>
auto vec(const int (&d)[n],const T &init=T()){
  if constexpr (id<n)return vector(d[id],vec<T,n,id+1>(d,init));
  else return init;
}
#ifdef LOCAL
#include<debug.h>
#define SWITCH(a,b) (a)
#else
#define debug(...) static_cast<void>(0)
#define debugg(...) static_cast<void>(0)
#define SWITCH(a,b) (b)
template<typename T1,typename T2>ostream &operator<<(ostream &os,const pair<T1,T2>&p){os<<p.first<<' '<<p.second;return os;}
#endif
struct Timer{
  clock_t start;
  Timer(){
    start=clock();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout<<fixed<<setprecision(16);
  }
  inline double now(){return (double)(clock()-start)/1000;}
  #ifdef LOCAL
  ~Timer(){
    cerr<<"time:";
    cerr<<now();
    cerr<<"ms\n";
  }
  #endif
}timer;
void SOLVE();
int main(){
  int testcase=1;
  //cin>>testcase;
  for(int i=0;i<testcase;i++){
    SOLVE();
  }
}

template <typename T, T (*add)(T, T), T (*mul)(T, T), T (*I0)(), T (*I1)()>
struct semiring {
  T x;
  semiring() : x(I0()) {}
  semiring(T y) : x(y) {}
  static T id0() { return I0(); }
  static T id1() { return I1(); }

  semiring &operator+=(const semiring &p) {
    if (x == I0()) return *this = p;
    if (p.x == I0()) return *this;
    return *this = add(x, p.x);
  }

  semiring &operator*=(const semiring &p) {
    if (x == I0() || p.x == I0()) return *this = I0();
    if (x == I1()) return *this = p;
    if (p.x == I1()) return *this;
    return *this = mul(x, p.x);
  }

  semiring operator+(const semiring &p) const { return semiring(*this) += p; }
  semiring operator*(const semiring &p) const { return semiring(*this) *= p; }
  bool operator==(const semiring &p) const { return x == p.x; }
  bool operator!=(const semiring &p) const { return x != p.x; }
  friend ostream &operator<<(ostream &os, const semiring &p) {
    return os << p.x;
  }
};

template <typename rig>
struct Mat {
  using Array = vector<vector<rig>>;
  Array A;
  int N;

  Mat() {
    // for (int i = 0; i < N; i++) A[i].fill(rig::id0());
  }
  Mat(vector<vector<ll>>a){
    N=a.size();
    A.resize(N,vector<rig>(N));
    rep(i,a.size())rep(j,a.size())A[i][j]=a[i][j];
  }
  int height() const { return N; }
  int width() const { return N; }
  inline const vector<rig> &operator[](int k) const { return A[k]; }
  inline vector<rig> &operator[](int k) { return A[k]; }

  static Mat I(int N) {
    Mat m;
    m.A=vector<vector<rig>>(N,vector<rig>(N));
    m.N=N;
    for (int i = 0; i < N; i++) m[i][i] = rig::id1();
    return m;
  }

  Mat &operator+=(const Mat &B) {
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++) A[i][j] += B[i][j];
    return (*this);
  }

  Mat &operator*=(const Mat &B) {
    Mat C;
    C.A=vector<vector<rig>>(N,vector<rig>(N));
    C.N=N;
    for (int i = 0; i < N; i++){

      for (int k = 0; k < N; k++){

        for (int j = 0; j < N; j++){
          C[i][j] += A[i][k] * B[k][j];
        }
      }
    }
    swap(A,C.A);
    return (*this);
  }

  Mat &operator^=(long long k) {
    Mat B = Mat::I(N);
    for (; k; *this *= *this, k >>= 1){
      debug('U');
      if (k & 1) B *= *this;
      debug('U');
    }
    A.swap(B.A);
    return (*this);
  }

  Mat operator+(const Mat &B) const { return (Mat(*this) += B); }
  Mat operator*(const Mat &B) const { return (Mat(*this) *= B); }
  Mat operator^(long long k) const { return (Mat(*this) ^= k); }

  friend ostream &operator<<(ostream &os, Mat &p) {
    int N=p.A.size();
    for (int i = 0; i < N; i++) {
      os << "[";
      for (int j = 0; j < N; j++) {
        os << p[i][j].x << (j == N - 1 ? "]\n" : ",");
      }
    }
    return (os);
  }
};
constexpr ll infLL=1e18;
using U = long long;
U add(U a, U b) { return max(a, b); }
U mul(U a, U b) { return a + b; }
U i0() { return -infLL; }
U i1() { return 0; }
using rig = semiring<U, add, mul, i0, i1>;
//https://nyaannyaan.github.io/library/math/semiring.hpp.html
void SOLVE(){
  int n;
  ll m;
  cin>>n>>m;
  vector<ll>t(n);
  cin>>t;
  vector<vector<ll>>a(n,vector<ll>(n));
  cin>>a;
  Mat<rig> mat(a);
  mat^=m;
  rep(i,n){
    ll ans=i0();
    rep(j,n)ans=add(ans,mul(t[j],mat[j][i].x));

    cout<<ans<<" \n"[i+1==n];
  }
}
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