Submission #1129206

#	Time	Username	Problem	Language	Result	Execution time	Memory
1129206		RedGrey1993	trapezoid (balkan11_trapezoid)	C++20	43 / 100	338 ms	49316 KiB

trapezoid

#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

// For example: SegmentTree<int> st(n, [](int a, int b) { return max(a, b); });
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]);
    }
  }
 
  // 迭代写法，效率更高（多找几道题验证下）
  T Query(int lo, int hi) {
    T ra = default_value_, rb = default_value_;
    for (lo += offset_ - 1, hi += offset_ - 1; lo < hi; lo /= 2, hi /= 2) {
      if (!(lo & 1)) ra = f_(ra, nodes_[lo++]);
      if (!(hi & 1)) rb = f_(rb, nodes_[--hi]);
    }
    return f_(ra, rb);
  }
  // 递归写法，更好理解
  // T Query(int a, int b) { return Query(a, b, 0, 0, offset_); }
 
 private:
  // 求[a,b)区间的目标值
  // 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);
  T Query(int a, int b, int k, int l, int r) {
    // 如果[a,b)和[l,r)不相交，返回0
    if (r <= a || b <= l) return default_value_;
    // 如果[a,b)完全包含[l,r)，返回当前节点值
    if (a <= l && r <= b)
      return nodes_[k];
    else {
      T v1 = Query(a, b, k * 2 + 1, l, (l + r) / 2);
      T v2 = Query(a, b, k * 2 + 2, (l + r) / 2, r);
      return f_(v1, v2);
    }
  }
  int node_num_;
  int offset_;
  vector<T> nodes_;
  function<T(T, T)> f_;
  T default_value_;
};

constexpr int MOD = 30013;

class Solution {
    struct Line {
        int up,down;
        bool is_left;
        int id;
        pii max_independent_set = mp(0, 0);
        friend ostream& operator<<(ostream& os, const Line& line) {
            os << "[" << (line.is_left ? "L" : "R") << "](" << line.up << "," << line.down << ")";
            return os;
        }
        bool operator<(const Line& other) const {
            return down == other.down ? up < other.up : down < other.down;
        }
    };
public:
    void Solve() {
        int n;
        while(cin>>n) {
            // 存储梯形的坐标信息
            // 每个梯形由两个点对表示:{{左上,左下}, {右上,右下}}
            vector<Line> trapezoids;
            set<int> coordinates;
            // 读入n个梯形的坐标
            for (int i = 0; i < n; i++) {
                int upper_left, upper_right;    // 上边两端点的x坐标
                int bottom_left, bottom_right;  // 下边两端点的x坐标
                
                cin >> upper_left >> upper_right >> bottom_left >> bottom_right;
                coordinates.insert(upper_left);
                coordinates.insert(upper_right);
                coordinates.insert(bottom_left);
                coordinates.insert(bottom_right);
                
                trapezoids.push_back({upper_left, bottom_left, true, i});
                trapezoids.push_back({upper_right, bottom_right, false, i});
            }

            map<int, int> cordinate_compress;
            int i = 0;
            for (auto x : coordinates) {
                cordinate_compress[x] = i++;
            }
            
            sort(trapezoids.begin(), trapezoids.end());
            map<int, int> right_line_id_to_index;
            for(int i = 0; i < trapezoids.size(); i++) {
                if(!trapezoids[i].is_left) {
                    right_line_id_to_index[trapezoids[i].id] = i;
                }
            }
            
            SegmentTree<pii> st(coordinates.size(), [](pii a, pii b) { 
                if (a.first > b.first) return a;
                else if (a.first < b.first) return b;
                else return mp(a.first, (a.second + b.second) % MOD);
            }, mp(0, 0));
            pii ans = mp(0, 0);
            for(int i = 0; i < trapezoids.size(); i++) {
                if(trapezoids[i].is_left) {
                    pii ret = st.Query(0, cordinate_compress[trapezoids[i].up] + 1);
                    if (ret.second == 0) ret.second = 1;
                    ret.first = (ret.first + 1) % MOD;
                    auto& right_line = trapezoids[right_line_id_to_index[trapezoids[i].id]];
                    right_line.max_independent_set = ret;
                    ans = max(ans, ret);
                } else {
                    auto& right_line = trapezoids[i];
                    st.Update(cordinate_compress[right_line.up], right_line.max_independent_set);
                }
            }

            // 输出所有梯形的坐标信息
            cout << ans.first << " " << ans.second << endl;
        }
        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: 43 / 100