結果
| 問題 |
No.2446 完全列
|
| コンテスト | |
| ユーザー |
risujiroh
|
| 提出日時 | 2023-08-25 22:31:26 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 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 | 3,264 ms |
| コンパイル使用メモリ | 268,404 KB |
| 実行使用メモリ | 5,248 KB |
| 最終ジャッジ日時 | 2024-12-24 14:21:57 |
| 合計ジャッジ時間 | 4,072 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 30 |
コンパイルメッセージ
In file included from main.cpp:7:
main.cpp:53: warning: "assert" redefined
53 | #define assert(expr) (expr) || (__builtin_unreachable(), 0)
|
In file included from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/cassert:44,
from /home/linuxbrew/.linuxbrew/Cellar/gcc@12/12.3.0/include/c++/12/x86_64-pc-linux-gnu/bits/stdc++.h:33,
from main.cpp:3:
/usr/include/assert.h:92: note: this is the location of the previous definition
92 | # 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__
risujiroh