結果

問題 No.704 ゴミ拾い Medium
ユーザー fumiphysfumiphys
提出日時 2019-07-18 00:55:56
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 236 ms / 1,500 ms
コード長 5,515 bytes
コンパイル時間 1,884 ms
コンパイル使用メモリ 180,536 KB
実行使用メモリ 29,056 KB
最終ジャッジ日時 2024-12-23 17:41:31
合計ジャッジ時間 10,395 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 44
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>

// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#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 vrep(v, i) for(int i = 0; i < (v).size(); i++)
#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 FI first
#define SE second
#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

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
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef complex<double> cd;

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
template<typename T, typename E>
struct SegmentTree_ {
  function<T(T, T)> f;
  function<T(T, E)> g;
  int n;
  T def;
  vector<T> vec;
  SegmentTree_(){}
  SegmentTree_(int n_, function<T(T, T)> f_, function<T(T, E)> g_, T def_, vector<T> v=vector<T>()){
    f = f_;
    g = g_;
    def = def_;

    // initialize vector
    n = 1;
    while(n < n_){
      n *= 2;
    }
    vec = vector<T>(2*n -1, def);

    // initialize segment tree
    for(int i = 0; i < v.size(); i++){
      vec[i + n - 1] = v[i];
    }
    for(int i = n - 2; i >= 0; i--){
      vec[i] = f(vec[2*i+1], vec[2*i+2]);
    }
  }
  void update(int k, const E &val){
    k = k + n - 1;
    vec[k] = g(vec[k], val);
    while(k > 0){
      k = (k - 1) / 2;
      vec[k] = f(vec[2*k+1], vec[2*k+2]);
    }
  }
  // [l, r) -> [a, b) (at k)
  T query(int a, int b, int k, int l, int r){
    if(r <= a || b <= l)return def;
    if(a <= l && r <= b)return vec[k];
    T ld = query(a, b, 2*k+1, l, (l+r)/2);
    T rd = query(a, b, 2*k+2, (l+r)/2, r);
    return f(ld, rd);
  }
  T query(int a, int b){
    return query(a, b, 0, 0, n);
  }
};

template<typename T, typename E>
using SegmentTree = struct SegmentTree_<T, E>;
using SegmentTreeI = SegmentTree<int, int>;
using SegmentTreeL = SegmentTree<long long, long long>;
ll dp[300010];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); vector<ll> a(n), x(n), y(n); cin >> a >> x >> y;
  SegmentTreeL segp = SegmentTreeL(n, [](ll a, ll b){return min(a, b);},
                                   [](ll a,ll b){return b;}, linf, vector<ll>());
  SegmentTreeL segm = SegmentTreeL(n, [](ll a, ll b){return min(a, b);},
                                   [](ll a,ll b){return b;}, linf, vector<ll>());
  segp.update(0, y[0] + x[0]);
  segm.update(0, y[0] - x[0]);
  rep(i, n + 1)dp[i] = linf;
  dp[0] = 0;
  rep(i, n){
    int idx = lower_bound(all(x), a[i]) - x.begin();
    dp[i+1] = min(dp[i+1], segp.query(idx, n) - a[i]);
    dp[i+1] = min(dp[i+1], segm.query(0, idx) + a[i]);
    if(i + 1 < n){
      segp.update(i + 1, dp[i+1] + y[i+1] + x[i+1]);
      segm.update(i + 1, dp[i+1] + y[i+1] - x[i+1]);
    }
  }
  cout << dp[n] << endl;
  return 0;
}
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