Submission #1108875

#	Time	Username	Problem	Language	Result	Execution time	Memory
1108875		RedGrey1993	Deda (COCI17_deda)	C++17	140 / 140	96 ms	6984 KiB

deda

#include <bits/stdc++.h>
 
using namespace std;
 
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &pa) { is >> pa.first >> pa.second; return is; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << "(" << pa.first << "," << pa.second << ")"; return os; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template <typename T> void resize_array(vector<T> &vec, int len) { vec.resize(len); }
template <typename T, typename... Args> void resize_array(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) resize_array(v, args...); }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }

// 函数名后面加 ll 就可以计算 long long类型对应的结果
// __builtin_ffs(x)  // 返回x中最后一个为1的位是从后向前的第几位
// __builtin_popcount(x)  // x中1的个数
// __builtin_ctz(x)  // x末尾0的个数。x=0时结果未定义。
// __builtin_clz(x)  // x前导0的个数。x=0时结果未定义。
// __builtin_parity(x)  // x中1的奇偶性。
#define highest_bit1_index(x) (31 - __builtin_clz(x))
#define highest_bit1_index_ll(x) (63 - __builtin_clzll(x))
#define rep(i, a, n) for (int i = a; i < (n); i++)
#define per(i, a, n) for (int i = (n)-1; i >= a; i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
typedef vector<int> vi;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef double db;
#if DEBUG
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#else
#define dbg(x)
#endif


template <typename T>
class SegmentTree {
 public:
  SegmentTree(int node_num, function<T(T, T)> f, T default_v = T())
      : f_(f), default_value_(default_v) {
    node_num_ = node_num;
    offset_ = 1;
    while (offset_ < node_num) offset_ *= 2;
    nodes_.resize(offset_ * 2, default_v);
  }

  friend istream& operator>>(istream& in, SegmentTree<T>& st) {
    for (int i = 0; i < st.node_num_; ++i) {
        cin >> st.nodes_[i + st.offset_ - 1];
    }
    return in;
  }

  // Init 前需要先赋初始值给nodes_
  // rep(i,0,n) {
  //     nodes_[i + offset_ - 1] = A[i];
  // }
  void Init() {
    for (int k = offset_ - 2; k >= 0; --k) {
        nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]);
    }
  }

  // 将第k个值更新为a
  void Update(int k, T a) {
    // 叶子节点
    k += offset_ - 1;
    nodes_[k] = a;
    // 向上更新
    while (k > 0) {
      k = (k - 1) / 2;
      nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]);
    }
  }
 
  T Query(int a, int b, int x) { return Query(a, b, 0, 0, offset_, x); }
 
 private:
  // 求[a,b)区间第一个值<=x的下标, f = min(a,b)
  // 后面的参数是为了计算方便传入的
  // k是节点的编号，l,r表示这个节点代表区间[l,r)
  // 外部调用时，使用query(a,b,0,0,n);
  int Query(int a, int b, int k, int l, int r, int x) {
    // 如果[a,b)和[l,r)不相交，返回0
    if (r <= a || b <= l) return -1;
    // 如果[a,b)完全包含[l,r)
    if (a <= l && r <= b) {
      if (x < nodes_[k]) {
          return -1;
      }
      if (l + 1 == r) {return l;}
      int i = Query(a, b, k * 2 + 1, l, (l + r) / 2, x);
      if (i != -1) return i;
      return Query(a, b, k * 2 + 2, (l + r) / 2, r, x);
    } else {
      int v1 = Query(a, b, k * 2 + 1, l, (l + r) / 2, x);
      if (v1 != -1) return v1;
      return Query(a, b, k * 2 + 2, (l + r) / 2, r, x);
    }
  }
  int offset_;
  int node_num_;
  vector<T> nodes_;
  function<T(T, T)> f_;
  T default_value_;
};

class Solution {
public:
    void Solve() {
        int n,m;
        while(cin>>n>>m) {
            SegmentTree<int> st(n, [](int a, int b) {
                return min(a,b);
            }, numeric_limits<int>::max());

            string op;
            int x,y;
            rep(i,0,m) {
                cin>>op>>x>>y;
                if (op == "M") {
                    st.Update(y-1, x);
                } else {
                    int idx = st.Query(y-1, n, x);
                    if (idx >=0) idx++;
                    cout << idx << '\n';
                }
            }
        }
        cout.flush();
    }
private:
};

// # define FILE_IO 1
void set_io(const string &name = "") {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
#if FILE_IO
    if (!name.empty()) {
        freopen((name + ".in").c_str(), "r", stdin);
        freopen((name + ".out").c_str(), "w", stdout);
    }
#endif
}

int main() {
    set_io("tmp");
    Solution().Solve();

    return 0;
}

• Subtask 1: 140 / 140