#907907 (C++23) No.2446 完全列

• 問題: No.2446 完全列
• ユーザー: risujiroh
• 提出日時: 2023-08-25 22:31:26
• 言語: C++23 (gcc 12.3.0 + boost 1.83.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 4,618 bytes
• コンパイル時間: 3,205 ms
• コンパイル使用メモリ: 265,756 KB
• 実行使用メモリ: 4,380 KB
• 最終ジャッジ日時: 2023-08-26 01:34:41
• 合計ジャッジ時間: 4,380 ms
• ジャッジサーバーID （参考情報）: judge13 / judge15

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	2 ms; 4,356 KB
testcase_01	AC	2 ms; 4,380 KB
testcase_02	AC	2 ms; 4,376 KB
testcase_03	AC	2 ms; 4,380 KB
testcase_04	AC	2 ms; 4,380 KB
testcase_05	AC	1 ms; 4,380 KB
testcase_06	AC	1 ms; 4,376 KB
testcase_07	AC	1 ms; 4,376 KB
testcase_08	AC	2 ms; 4,380 KB
testcase_09	AC	2 ms; 4,380 KB
testcase_10	AC	1 ms; 4,376 KB
testcase_11	AC	1 ms; 4,380 KB
testcase_12	AC	1 ms; 4,376 KB
testcase_13	AC	2 ms; 4,380 KB
testcase_14	AC	2 ms; 4,380 KB
testcase_15	AC	1 ms; 4,376 KB
testcase_16	AC	2 ms; 4,376 KB
testcase_17	AC	2 ms; 4,380 KB
testcase_18	AC	2 ms; 4,380 KB
testcase_19	AC	1 ms; 4,376 KB
testcase_20	AC	1 ms; 4,380 KB
testcase_21	AC	2 ms; 4,376 KB
testcase_22	AC	2 ms; 4,380 KB
testcase_23	AC	2 ms; 4,376 KB
testcase_24	AC	2 ms; 4,376 KB
testcase_25	AC	1 ms; 4,376 KB
testcase_26	AC	2 ms; 4,376 KB
testcase_27	AC	2 ms; 4,380 KB
testcase_28	AC	1 ms; 4,380 KB
testcase_29	AC	1 ms; 4,376 KB

コンパイルメッセージ

次のファイルから読み込み:  main.cpp:7:
main.cpp:53: 警告: "assert" が再定義されました
   53 | #define assert(expr) (expr) || (__builtin_unreachable(), 0)
      | 
次のファイルから読み込み:  /usr/local/gcc7/include/c++/12.2.0/cassert:44,
         次から読み込み:  /usr/local/gcc7/include/c++/12.2.0/x86_64-pc-linux-gnu/bits/stdc++.h:33,
         次から読み込み:  main.cpp:3:
/usr/include/assert.h:89: 備考: ここが以前の宣言がある位置です
   89 | #  define assert(expr)                                                  \
      | 

ソースコード

#if __INCLUDE_LEVEL__ == 0

#include <bits/stdc++.h>

using namespace std;

#include __BASE_FILE__

namespace {

int get_rank(vector<vector<double>> a) {
  vector<double> b(a.size());
  vector<double> x(a[0].size());
  return kactl::solveLinear(a, b, x);
}

void solve() {
  int l, m, n;
  cin >> tie(l, m, n);
  vector a(l, vector<double>(m));
  vector b(m, vector<double>(n));
  cin >> tie(a, b);
  vector c(l, vector<double>(n));
  for (int i : rep(l)) {
    for (int k : rep(m)) {
      for (int j : rep(n)) {
        c[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  for (int i : rep(l)) {
    for (int j : rep(n)) {
      if (kactl::eps < abs(c[i][j])) {
        print("No");
        return;
      }
    }
  }
  print(m - get_rank(a) == get_rank(b) ? "Yes" : "No");
}

}  // namespace

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  solve();
}

#else  // __INCLUDE_LEVEL__

#define assert(expr) (expr) || (__builtin_unreachable(), 0)

template <class T, class U = T>
bool chmin(T& x, U&& y) {
  return y < x && (x = forward<U>(y), true);
}

template <class T, class U = T>
bool chmax(T& x, U&& y) {
  return x < y && (x = forward<U>(y), true);
}

namespace std {

template <class T1, class T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
  return is >> p.first >> p.second;
}

template <class... Ts>
istream& operator>>(istream& is, tuple<Ts...>& t) {
  return apply([&is](auto&... xs) -> istream& { return (is >> ... >> xs); }, t);
}

template <class... Ts>
istream& operator>>(istream& is, tuple<Ts&...>&& t) {
  return is >> t;
}

template <class R, enable_if_t<!is_convertible_v<R, string>>* = nullptr>
auto operator>>(istream& is, R&& r) -> decltype(is >> *begin(r)) {
  for (auto&& e : r) {
    is >> e;
  }
  return is;
}

template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << p.first << ' ' << p.second;
}

template <class... Ts>
ostream& operator<<(ostream& os, const tuple<Ts...>& t) {
  auto f = [&os](const auto&... xs) -> ostream& {
    [[maybe_unused]] auto sep = "";
    ((os << exchange(sep, " ") << xs), ...);
    return os;
  };
  return apply(f, t);
}

template <class R, enable_if_t<!is_convertible_v<R, string_view>>* = nullptr>
auto operator<<(ostream& os, R&& r) -> decltype(os << *begin(r)) {
  auto sep = "";
  for (auto&& e : r) {
    os << exchange(sep, " ") << e;
  }
  return os;
}

}  // namespace std

template <class... Ts>
void print(Ts&&... xs) {
  cout << tie(xs...) << '\n';
}

inline auto rep(int l, int r) { return views::iota(min(l, r), r); }
inline auto rep(int n) { return rep(0, n); }
inline auto rep1(int l, int r) { return rep(l, r + 1); }
inline auto rep1(int n) { return rep(1, n + 1); }
inline auto per(int l, int r) { return rep(l, r) | views::reverse; }
inline auto per(int n) { return per(0, n); }
inline auto per1(int l, int r) { return per(l, r + 1); }
inline auto per1(int n) { return per(1, n + 1); }

// https://github.com/kth-competitive-programming/kactl
namespace kactl {

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

/**
 * Author: Per Austrin, Simon Lindholm
 * Date: 2004-02-08
 * License: CC0
 * Description: Solves $A * x = b$. If there are multiple solutions, an
 * arbitrary one is returned. Returns rank, or -1 if no solutions. Data in $A$
 * and $b$ is lost. Time: O(n^2 m) Status: tested on kattis:equationsolver, and
 * bruteforce-tested mod 3 and 5 for n,m <= 3
 */

typedef vector<double> vd;
const double eps = 1e-12;

int solveLinear(vector<vd>& A, vd& b, vd& x) {
  int n = sz(A), m = sz(x), rank = 0, br, bc;
  if (n) assert(sz(A[0]) == m);
  vi col(m);
  iota(all(col), 0);

  rep(i, 0, n) {
    double v, bv = 0;
    rep(r, i, n) rep(c, i, m) if ((v = fabs(A[r][c])) > bv) br = r, bc = c,
                                                            bv = v;
    if (bv <= eps) {
      rep(j, i, n) if (fabs(b[j]) > eps) return -1;
      break;
    }
    swap(A[i], A[br]);
    swap(b[i], b[br]);
    swap(col[i], col[bc]);
    rep(j, 0, n) swap(A[j][i], A[j][bc]);
    bv = 1 / A[i][i];
    rep(j, i + 1, n) {
      double fac = A[j][i] * bv;
      b[j] -= fac * b[i];
      rep(k, i + 1, m) A[j][k] -= fac * A[i][k];
    }
    rank++;
  }

  x.assign(m, 0);
  for (int i = rank; i--;) {
    b[i] /= A[i][i];
    x[col[i]] = b[i];
    rep(j, 0, i) b[j] -= A[j][i] * b[i];
  }
  return rank;  // (multiple solutions if rank < m)
}

#undef sz
#undef all
#undef rep

}  // namespace kactl

#endif  // __INCLUDE_LEVEL__