結果

問題 No.270 next_permutation (1)
ユーザー fumiphysfumiphys
提出日時 2019-10-26 20:59:10
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 23 ms / 2,000 ms
コード長 3,755 bytes
コンパイル時間 1,418 ms
コンパイル使用メモリ 169,012 KB
実行使用メモリ 4,704 KB
最終ジャッジ日時 2023-10-12 19:36:24
合計ジャッジ時間 2,700 ms
ジャッジサーバーID
（参考情報） 		judge11 / judge12
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 2 ms
4,352 KB
testcase_01 AC 2 ms
4,348 KB
testcase_02 AC 2 ms
4,348 KB
testcase_03 AC 2 ms
4,352 KB
testcase_04 AC 2 ms
4,348 KB
testcase_05 AC 2 ms
4,352 KB
testcase_06 AC 2 ms
4,348 KB
testcase_07 AC 1 ms
4,352 KB
testcase_08 AC 2 ms
4,348 KB
testcase_09 AC 4 ms
4,348 KB
testcase_10 AC 22 ms
4,568 KB
testcase_11 AC 23 ms
4,696 KB
testcase_12 AC 23 ms
4,704 KB
testcase_13 AC 21 ms
4,616 KB
testcase_14 AC 22 ms
4,564 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 # 

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

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;


int main(int argc, char const* argv[])
{
  int n, k; cin >> n >> k;
  vector<ll> p(n), b(n);
  cin >> p >> b;
  if(k == 0){
    cout << 0 << endl;
    return 0;
  }
  ll res = 0;
  ll sum = 0;
  rep(i, n)sum += abs(p[i] - b[i]);
  res += sum;
  auto p_swap = [&](int i, int j){
    sum -= abs(p[i] - b[i]);
    sum -= abs(p[j] - b[j]);
    swap(p[i], p[j]);
    sum += abs(p[i] - b[i]);
    sum += abs(p[j] - b[j]);
  };
  rep(i_, k - 1){
    int i = n - 2;
    for(; i >= 0; i--){
      if(p[i] < p[i+1])break;
    }
    if(i == -1){
      rep(j, n)p[j] = j + 1;
      sum = 0;
      rep(j, n)sum += abs(p[j] - b[j]);
    }else{
      int j = n - 1;
      for(; j > i; j--){
        if(p[j] > p[i])break;
      }
      p_swap(i, j);
      for(int l = i + 1, m = n - 1; l < m; l++, m--){
        p_swap(l, m);
      }
    }
    res += sum;
  }
  cout << res << endl;
  return 0;
}
0