Submission #887837

#	Time	Username	Problem	Language	Result	Execution time	Memory
887837		ad_red	Tri (CEOI09_tri)	C++17	0 / 100	1020 ms	10404 KiB

tri

#include <bits/stdc++.h>
#define endl "\n"
  
using namespace std;
using ll = long long;

  
struct Point{
  ll x, y;
};
  
ll vp(Point a, Point b) {
  return a.x * b.y - a.y * b.x;
}
  
ll sgn(Point a, Point b, Point c) {
  // -1 if the order is A-B-C from left to right if B is the bottom point
  // 1 or 0 otherwise
  
  ll q = vp(Point{a.x - b.x, a.y - b.y}, Point{c.x - b.x, c.y - b.y});
  return (q / abs(q));
}
  
bool operator<(Point a, Point b) {
  return sgn(a, Point{0LL, 0LL}, b) == -1;
}
  
bool in_triangle(Point a, Point b, Point c, Point p) {
  // assuming A-B-C
  
  return (sgn(a, b, p) == -1 && sgn(c, b, p) == 1 && sgn(p, c, a) == -1);
}
  
bool cmp_hull(Point a, Point b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
  
/*
  Plan:
  0. Sort all points by angle
  1. Construct sqrt(n) convex hulls for all point sets
  2. For each triangle, consider all sqrt(n) ranges of points already present
  3. Check all points that are outside of the hulls manually
  3.5 On both sides
  4. For each complete range with a hull do a binary search on that hull:
  5. Start with the leftmost (by the angle) point, end with the point anticlockwise on the convex hull
  6. Check if the mid is in the triangle, if it is, then break. If we are moving further from the triangle by choosing a point to the right of the current one (cur_mid), then r = mid, else l = mid.
  
  Claim: the total thing takes no more than 200 lines.
*/
  
vector<Point> points;
const ll sqrt_size = 600;

signed main() {
  ll n, m;
  cin >> n >> m;
  
  for (ll i = 0; i < n; i++) {
    ll x, y;
    cin >> x >> y;
  
    points.push_back(Point{x, y});
  }
  
  sort(points.begin(), points.end()); // the comparator is there
  
  vector<vector<Point>> hulls(n);
  
  for (ll i = 0; i < n; i++) {
    hulls[i / sqrt_size].push_back(points[i]);
  }
  
  
  for (ll i = 0; i < n; i++) {
    if (hulls[i].empty()) continue;
    
    vector<Point> hull;
  
    sort(hulls[i].begin(), hulls[i].end(), cmp_hull);
  
    for (auto p : hulls[i]) {
      while (hull.size() > 1 && sgn(p, hull[(ll)hull.size() - 2], hull.back()) == -1) {
        hull.pop_back();
      }
      hull.push_back(p);
    }
  
    hulls[i] = hull;
    // top convex hull only!
  }
  
  // end of hull processing
  
  for (ll trn = 0; trn < m; trn++) {
    // current triangle
  
    Point a, b;
    cin >> a.x >> a.y >> b.x >> b.y;
  
    if (sgn(a, Point{0LL, 0LL}, b) >= 0) swap(a, b);
  
    ll left_start = n - 1, right_end = 0;
  
    // left_start - leftmost point that is in the angle
    // right_end - leftmost point after the angle
    {
      ll l = -1;
      ll r = n - 1;
    
      while (r - l > 1) {
        ll mid = (l + r) / 2;
        if (sgn(a, Point{0, 0}, points[mid]) >= 0) {
          l = mid;
        } else {
          r = mid;
        }
      }
    
      left_start = r;
    }
  
    {
      ll l = 0;
      ll r = n;
    
      while (r - l > 1) {
        ll mid = (l + r) / 2;
        if (sgn(b, Point{0, 0}, points[mid]) >= 0) {
          l = mid;
        } else {
          r = mid;
        }
      }
    
      right_end = r;
    }
  
    if (left_start >= right_end) {
      cout << "N" << endl;
      continue;
    }
  
    bool flag = false;
  
    if (right_end - left_start <= sqrt_size) {
      for (ll i = left_start; i < right_end; i++) {
        if (in_triangle(a, Point{0, 0}, b, points[i])) {
          flag = true;
          break;
        }
      }
  
      if (flag) {
        cout << "Y" << endl;
      } else {
        cout << "N" << endl;
      }
  
      continue;
    }
  
    flag = false;
  
    while (left_start % sqrt_size != 0) {
      if (in_triangle(a, Point{0, 0}, b, points[left_start])) {
        flag = true;
      }
      left_start++;
    }
  
    while (right_end - 1 > left_start && right_end % sqrt_size != 0) {
      right_end--;
      if (in_triangle(a, Point{0, 0}, b, points[right_end])) {
        flag = true;
      }
    }
  
    flag = false;
  
    for (ll i = left_start / sqrt_size; i < right_end / sqrt_size; i++) {
      // convex hull processing
  
      ll l = 0;
      ll r = (ll)hulls[i].size();
  
      while (r - l > 1) {
        ll mid = (r + l) / 2;
  
        if (in_triangle(a, Point{0, 0}, b, hulls[i][mid])) {
          flag = true;
          break;
        }
  
        if (mid + 1 == hulls[i].size() || !in_triangle(a, Point{0, 0}, b, hulls[i][mid + 1]) && sgn(hulls[i][mid + 1], a, hulls[i][mid]) == -1) {
          r = mid;
        } else {
          l = mid;
        }
      }
  
      for (ll j = (l - 3 + hulls[i].size()) % hulls[i].size(); j <= (r + 3 + hulls[i].size()) % hulls[i].size(); j++) {
        if (in_triangle(a, Point{0, 0}, b, hulls[i][j])) {
          flag = true;
          break;
        }
      }
  
      if (flag) break;
    }
  
    if (flag) {
      cout << "Y" << endl;
    } else {
      cout << "N" << endl;
    }
  }
  
  return 0;
}

Compilation message (stderr)

tri.cpp: In function 'int main()':
tri.cpp:196:21: warning: comparison of integer expressions of different signedness: 'll' {aka 'long long int'} and 'std::vector<Point>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  196 |         if (mid + 1 == hulls[i].size() || !in_triangle(a, Point{0, 0}, b, hulls[i][mid + 1]) && sgn(hulls[i][mid + 1], a, hulls[i][mid]) == -1) {
      |             ~~~~~~~~^~~~~~~~~~~~~~~~~~
tri.cpp:196:94: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
  196 |         if (mid + 1 == hulls[i].size() || !in_triangle(a, Point{0, 0}, b, hulls[i][mid + 1]) && sgn(hulls[i][mid + 1], a, hulls[i][mid]) == -1) {
      |                                           ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
tri.cpp:203:66: warning: comparison of integer expressions of different signedness: 'll' {aka 'long long int'} and 'long long unsigned int' [-Wsign-compare]
  203 |       for (ll j = (l - 3 + hulls[i].size()) % hulls[i].size(); j <= (r + 3 + hulls[i].size()) % hulls[i].size(); j++) {
      |                                                                ~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 100