結果

問題 No.209 Longest Mountain Subsequence
ユーザー fumiphysfumiphys
提出日時 2019-10-05 14:14:43
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 61 ms / 2,000 ms
コード長 4,219 bytes
コンパイル時間 1,761 ms
コンパイル使用メモリ 167,964 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-04-15 18:38:41
合計ジャッジ時間 2,498 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ（β）

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 21 ms
6,816 KB
testcase_01 AC 20 ms
6,940 KB
testcase_02 AC 19 ms
6,940 KB
testcase_03 AC 61 ms
6,940 KB
testcase_04 AC 60 ms
6,940 KB
testcase_05 AC 6 ms
6,944 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 dpl2[101][101], dpr2[101][101];
int dpl[101], dpr[101];

int main(int argc, char const* argv[])
{
  int t; cin >> t;
  rep(i_, t){
    int n; cin >> n;
    vector<ll> a(n); cin >> a;
    rep(i, n + 1){
      dpl[i] = dpr[i] = 0;
      rep(j, n + 1){
        dpl2[i][j] = dpr2[i][j] = 0;
      }
    }
    for(int i = 0; i < n; i++){
      for(int j = 0; j < i; j++){
        if(a[j] >= a[i])continue;
        dpl2[j][i] = 2;
        for(int k = 0; k < j; k++){
          if(a[k] < a[j] && abs(a[k] - a[j]) < abs(a[j] - a[i])){
            dpl2[j][i] = max(dpl2[j][i], dpl2[k][j] + 1);
          }
        }
      }
    }
    rep(i, n)dpl[i] = 1;
    rep(i, n){
      for(int j = 0; j < i; j++){
        dpl[i] = max(dpl[i], dpl2[j][i]);
      }
    }
    for(int i = n - 1; i >= 0; i--){
      for(int j = n - 1; j > i; j--){
        if(a[i] <= a[j])continue;
        dpr2[i][j] = 2;
        for(int k = n - 1; k > j; k--){
          if(a[j] > a[k] && abs(a[j] - a[k]) < abs(a[i] - a[j])){
            dpr2[i][j] = max(dpr2[i][j], dpr2[j][k] + 1);
          }
        }
      }
    }
    rep(i, n)dpr[i] = 1;
    rep(i, n){
      for(int j = i + 1; j < n; j++){
        dpr[i] = max(dpr[i], dpr2[i][j]);
      }
    }
    int res = 0;
    rep(i, n){
      res = max(res, dpl[i] + dpr[i] - 1);
    }
    cout << res << endl;
  }
  return 0;
}
0