#381018 (C++14) No.151 セグメントフィッシング

提出ソース

結果

問題	No.151 セグメントフィッシング
ユーザー	fumiphys
提出日時	2019-09-20 20:50:13
言語	C++14 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	115 ms / 5,000 ms
コード長	5,470 bytes
コンパイル時間	1,922 ms
コンパイル使用メモリ	183,244 KB
実行使用メモリ	19,112 KB
最終ジャッジ日時	2023-10-12 16:37:15
合計ジャッジ時間	4,376 ms
ジャッジサーバーID （参考情報）	judge12 / judge15

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 4,348 KB
testcase_01	AC	2 ms 4,352 KB
testcase_02	AC	2 ms 4,348 KB
testcase_03	AC	2 ms 4,348 KB
testcase_04	AC	1 ms 4,348 KB
testcase_05	AC	2 ms 4,348 KB
testcase_06	AC	1 ms 4,348 KB
testcase_07	AC	2 ms 4,348 KB
testcase_08	AC	7 ms 4,352 KB
testcase_09	AC	7 ms 4,348 KB
testcase_10	AC	8 ms 4,348 KB
testcase_11	AC	7 ms 4,352 KB
testcase_12	AC	27 ms 6,176 KB
testcase_13	AC	28 ms 6,220 KB
testcase_14	AC	28 ms 6,408 KB
testcase_15	AC	28 ms 6,164 KB
testcase_16	AC	28 ms 6,220 KB
testcase_17	AC	9 ms 5,952 KB
testcase_18	AC	9 ms 5,852 KB
testcase_19	AC	49 ms 6,168 KB
testcase_20	AC	49 ms 6,228 KB
testcase_21	AC	16 ms 6,524 KB
testcase_22	AC	16 ms 6,504 KB
testcase_23	AC	115 ms 19,112 KB

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ソースコード

raw source code

// 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_;

template<typename T, typename E>
struct SegmentTree_ {
  function<T(T, T)> f; // aggregate function
  function<T(T, E)> g; // update function
  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>;


int main(int argc, char const* argv[])
{
  int n, q;
  cin >> n >> q;
  SegmentTreeL segl = SegmentTreeL(n + q + 1, [](ll a, ll b){return a + b;},
                                  [](ll a, ll b){return a + b;}, 0, vector<ll>());
  SegmentTreeL segr = SegmentTreeL(n + q + 1, [](ll a, ll b){return a + b;},
                                   [](ll a, ll b){return a + b;}, 0, vector<ll>());
  vector<vector<pair<int, ll>>> l(q), r(q);
  rep(i, q){
    char c;
    cin >> c;
    ll y, z;
    cin >> y >> z;
    for(auto e: l[i]){
      segl.update(e.first, - e.second);
      segr.update(q - i, e.second);
      if(n - 1 + i + 1 < q){
        r[n - 1 + i + 1].pb(mk(q - i, e.second));
      }
    }
    for(auto e: r[i]){
      segr.update(e.first, - e.second);
      segl.update(n - 1 + i, e.second);
      if(n - 1 + i + 1 < q){
        l[n - 1 + i + 1].pb(mk(n - 1 + i, e.second));
      }
    }
    if(c == 'L'){
      segl .update(y + i, z);
      if(y + i + 1 < q){
        l[y + i + 1].pb(mk(y + i, z));
      }
    }else if(c == 'R'){
      segr.update(y - i + q, z);
      if((n - 1) - y + i + 1 < q){
        r[(n - 1) - y + i + 1].pb(mk(y - i + q, z));
      }
    }else{
      cout << segl.query(y + i, z + i) + segr.query(y - i + q, z - i + q) << endl;
    }
  }
  return 0;
}

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結果

テストケース

ソースコード