結果

問題 No.1560 majority x majority
ユーザー stoqstoq
提出日時 2021-06-25 22:34:21
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 404 ms / 2,000 ms
コード長 5,032 bytes
コンパイル時間 2,194 ms
コンパイル使用メモリ 209,796 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-25 08:38:25
合計ジャッジ時間 5,368 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 404 ms
6,816 KB
testcase_01 AC 262 ms
6,940 KB
testcase_02 AC 2 ms
6,940 KB
testcase_03 AC 114 ms
6,940 KB
testcase_04 AC 78 ms
6,940 KB
testcase_05 AC 111 ms
6,940 KB
testcase_06 AC 27 ms
6,944 KB
testcase_07 AC 9 ms
6,940 KB
testcase_08 AC 125 ms
6,940 KB
testcase_09 AC 54 ms
6,940 KB
testcase_10 AC 2 ms
6,940 KB
testcase_11 AC 10 ms
6,940 KB
testcase_12 AC 4 ms
6,940 KB
testcase_13 AC 11 ms
6,940 KB
testcase_14 AC 4 ms
6,940 KB
testcase_15 AC 42 ms
6,940 KB
testcase_16 AC 2 ms
6,940 KB
testcase_17 AC 152 ms
6,940 KB
testcase_18 AC 31 ms
6,944 KB
testcase_19 AC 2 ms
6,940 KB
testcase_20 AC 3 ms
6,940 KB
testcase_21 AC 3 ms
6,940 KB
testcase_22 AC 62 ms
6,944 KB
testcase_23 AC 3 ms
6,940 KB
testcase_24 AC 2 ms
6,940 KB
testcase_25 AC 53 ms
6,940 KB
testcase_26 AC 2 ms
6,940 KB
testcase_27 AC 1 ms
6,944 KB
testcase_28 AC 2 ms
6,944 KB
testcase_29 AC 2 ms
6,940 KB
testcase_30 AC 2 ms
6,940 KB
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 #

#define MOD_TYPE 1

#pragma region Macros

#include <bits/stdc++.h>
using namespace std;

#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif

#if 0
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename T>
using smaller_queue = priority_queue<T, vector<T>, greater<T>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";

struct io_init
{
  io_init()
  {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b)
{
  if (a > b)
  {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b)
{
  if (a < b)
  {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b)
{
  return (a + b - 1) / b;
}
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val)
{
  fill((T *)array, (T *)(array + N), val);
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept
{
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept
{
  os << p.first << " " << p.second;
  return os;
}
#pragma endregion

// --------------------------------------

template <typename T>
struct SegmentTree
{
  using F = function<T(T, T)>;
  int N;
  vector<T> dat;
  T ti;
  F f;

  SegmentTree() {}
  SegmentTree(int n, F f, T ti) : f(f), ti(ti)
  {
    N = 1;
    while (N < n)
      N *= 2;
    dat.assign(2 * N - 1, ti);
  }
  SegmentTree(const vector<T> &v, F f, T ti) : f(f), ti(ti)
  {
    N = 1;
    int sz = v.size();
    while (N < sz)
      N *= 2;
    dat.resize(2 * N - 1);
    for (int i = 0; i < N; ++i)
      dat[i + N - 1] = (i < sz ? v[i] : ti);
    for (int i = N - 2; i >= 0; --i)
      dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);
  }

  void update(int k, T a)
  {
    k += N - 1;
    dat[k] = a;
    while (k > 0)
    {
      k = (k - 1) / 2;
      dat[k] = f(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }

  T get(int k) { return dat[k + N - 1]; }

  T query(int a, int b, int k, int l, int r)
  {
    if (r <= a or b <= l)
      return ti;
    if (a <= l and r <= b)
      return dat[k];
    T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return f(vl, vr);
  }
  T query(int a, int b) { return query(a, b, 0, 0, N); }
  T query_all() { return dat[0]; }

  void display()
  {
    for (int i = 0; i < N; i++)
      cerr << get(i) << " ";
    cerr << "\n";
  }
};

int n, m, q;

int E(vector<int> &L, vector<int> &R)
{
  int sum = 0;
  rep(i, n)
  {
    sum += abs(R[i] - L[i]);
  }
  return sum;
}

void solve()
{
  int n, m;
  cin >> n >> m;
  vector s(n, vector<int>(m));
  rep(i, n) rep(j, m) cin >> s[i][j];
  ll dp[1 << 12] = {};
  dp[0] = 1;
  repi(msk, 1 << m)
  {
    vector<int> people;
    repi(i, n)
    {
      bool b = true;
      repi(j, m)
      {
        if (!(msk & (1 << j)))
          continue;
        if (s[i][j] == 1)
          continue;
        b = false;
        break;
      }
      if (b)
        people.emplace_back(i);
    }
    rep(i, m)
    {
      if (msk & (1 << i))
        continue;
      int sz = people.size();
      int sum = 0;
      for (auto p : people)
      {
        sum += s[p][i];
      }
      if (sum * 2 >= sz)
      {
        dp[msk | (1 << i)] += dp[msk];
      }
    }
  }
  /*
  rep(i, 1 << m)
  {
    cout << i << " " << dp[i] << "\n";
  }*/
  cout << dp[(1 << m) - 1] << "\n";
}

int main()
{
  solve();
}
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