Submission #955983

#	Time	Username	Problem	Language	Result	Execution time	Memory
955983		chrisvilches	Dragon 2 (JOI17_dragon2)	C++14	100 / 100	2145 ms	4792 KiB

dragon2

#include <bits/stdc++.h>
using namespace std;
using ll = int;

ll Bx, By;

struct Point {
  ll x, y;
  int idx;
  inline Point operator-(const Point& p) const { return {x - p.x, y - p.y, idx}; }
  inline Point operator+(const Point& p) const { return {x + p.x, y + p.y, idx}; }
  inline long long cross(const Point& p) const {
    return x * (long long)p.y - y * (long long)p.x;
  }
  inline bool operator<(const Point& p) const {
    return to_upper().cross(p.to_upper()) > 0;
  }
  inline Point to_upper() const { return above() ? *this : negate(); }
  inline bool above() const {
    // TODO: Improve this
    // const Point B{Bx, By, idx};
    return Bx * (long long)y - By * (long long)x > 0;
    // return B.cross(*this) > 0;
  }
  inline Point negate() const { return {-x, -y, idx}; }
};

short orientation(const Point& o, const Point& a, const Point& b) {
  // assert((a - o).cross(b - o) != 0);
  const long long x = (a - o).cross(b - o);
  return (x > 0) - (x < 0);
}

short bit[30'001];
int bit_n = 30'001;

void clear(const int n) { memset(bit, 0, sizeof(short) * n); }

// TODO: Return should be int, not short.
int sum_single(int r) {
  int ret = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
  return ret;
}

int sum(int l, int r) { return sum_single(r) - sum_single(l - 1); }

void add(int idx, const short delta) {
  for (; idx < bit_n; idx = idx | (idx + 1)) bit[idx] += delta;
}

unordered_map<int, vector<Point>> order_by_b;

bool cmp_by_b(const Point& p, const Point& q) {
  const bool a1 = p.above();
  const bool a2 = q.above();
  if (a1 != a2) return a1;
  return p.cross(q) > 0;
}

int handle_query(const vector<Point>& points1, const vector<Point>& points2,
                 const vector<Point>& ord_b, const Point& B) {
  if (points1.empty() || points2.empty()) return 0;

  bit_n = (int)points2.size();
  clear(points2.size());

  for (const auto& q : points2) {
    if (!q.above()) {
      add(q.idx, 1);
    }
  }

  int total = 0;

  int j = 0;
  for (const Point& p : points1) {
    while (j < (int)points2.size()) {
      const Point& q = points2[j];

      if (!(q < p)) break;
      add(q.idx, q.above() ? 1 : -1);

      j++;
    }

    Point from_point = p;
    Point to_point = B + (p - B).negate();

    if (!p.above()) swap(from_point, to_point);

    // total += 1;
    // continue;

    const auto it1 = lower_bound(ord_b.begin(), ord_b.end(), from_point - B, cmp_by_b);
    const auto it2 = lower_bound(ord_b.begin(), ord_b.end(), to_point - B, cmp_by_b);

    const int from = it1 - ord_b.begin();
    const int to = it2 - ord_b.begin();

    total += sum(from, to - 1);
  }

  return total;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int N, M, Q;

  while (cin >> N >> M) {
    vector<vector<Point>> tribe_points(M + 1);
    for (int i = 0; i < N; i++) {
      Point p;
      int tribe;
      cin >> p.x >> p.y >> tribe;
      tribe_points[tribe].push_back(p);
    }

    Point A;
    cin >> A.x >> A.y;
    cin >> Bx >> By;
    cin >> Q;

    Bx -= A.x;
    By -= A.y;

    for (auto& points : tribe_points) {
      for (auto& p : points) p = p - A;
    }

    const Point B{Bx, By, -1};

    for (int m = 0; m <= M; m++) {
      auto& points = tribe_points[m];

      for (auto& p : tribe_points.at(m)) p = p - B;
      sort(points.begin(), points.end(), cmp_by_b);
      for (int i = 0; i < (int)points.size(); i++) {
        points[i].idx = i;
      }
      order_by_b[m] = tribe_points.at(m);
      for (auto& p : tribe_points.at(m)) p = p + B;
    }

    for (auto& points : tribe_points) {
      sort(points.begin(), points.end());
    }

    const Point origin{0, 0, -1};

    // TODO: Are the fenwick queries actually faster???? DO some experiments

    // TODO: Set this value more properly, and explain that this doesn't really help
    //       but it's something. (assuming fenwick is ACTUALLY faster)

    // TODO: Comment that this solution is probably not the intended one. The example
    // solution is much faster (under a second).
    // TODO: The thing is, I don't want to leave this code without the radial sweep, I
    // want to add it for the lulz.
    const int big_query = N / 2;


    while (Q--) {
      int i, j;
      cin >> i >> j;
      // cerr << tribe_points[i].size() * tribe_points[j].size() << endl;
      const long long size = tribe_points[i].size() * tribe_points[j].size();

      if (size > 100'000) {
        // cerr << "big" << endl;
        // cerr << tribe_points[j].size() << " > " << big_query << endl;
        const int ans =
            handle_query(tribe_points.at(i), tribe_points.at(j), order_by_b.at(j), B);
        cout << ans << '\n';
      } else {
        int total = 0;
        for (const Point& p : tribe_points[i]) {
          for (const Point& q : tribe_points[j]) {
            if (orientation(origin, B, p) == 1) {
              if (orientation(B, p, q) == 1 && orientation(p, origin, q) == 1) total++;
            } else {
              if (orientation(origin, p, q) == 1 && orientation(p, B, q) == 1) total++;
            }
          }
        }
        cout << total << '\n';
      }
    }
  
  }
}

Compilation message (stderr)

dragon2.cpp: In function 'int main()':
dragon2.cpp:162:15: warning: unused variable 'big_query' [-Wunused-variable]
  162 |     const int big_query = N / 2;
      |               ^~~~~~~~~

• Subtask 1: 15 / 15
• Subtask 2: 45 / 45
• Subtask 3: 40 / 40