답안 #888174

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
888174	2023-12-16T09:35:01 Z	ad_red	Tri (CEOI09_tri)	C++17	20 / 100	977 ms	6656 KB

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 = 1200;

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() >= 2 && sgn(p, hull[(ll)hull.size() - 2], hull.back()) == -1) {
        hull.pop_back();
      }
      hull.push_back(p);
    }

    hulls[i].clear();
    for (auto c : hull) {
      hulls[i].push_back(c);
    }
    // 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 in the angle
    // right_end - rightmost point in the angle
    {
      ll l = -1;
      ll r = n - 1;
    
      while (r - l > 1) {
        ll mid = (l + r) / 2;
        if (sgn(a, Point{0LL, 0LL}, 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{0LL, 0LL}, points[mid]) >= 0) {
          l = mid;
          right_end = l;
        } else {
          r = mid;
        }
      }
    }
  
    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{0LL, 0LL}, b, points[i])) {
          flag = true;
        }
      }
  
      if (flag) {
        cout << "Y" << endl;
      } else {
        cout << "N" << endl;
      }
  
      continue;
    }
  
    flag = false;
  
    while (left_start % sqrt_size != 0) {
      if (in_triangle(a, Point{0LL, 0LL}, b, points[left_start])) {
        flag = true;
      }
      left_start++;
    }
  
    while ((right_end >= left_start) && (right_end % sqrt_size != sqrt_size - 1)) {
      if (in_triangle(a, Point{0LL, 0LL}, b, points[right_end])) {
        flag = true;
      }
      right_end--;
    }

    assert(left_start % sqrt_size == 0);
    assert(right_end % sqrt_size == sqrt_size - 1);
  
    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() - 1;
  
      while (l <= r) {
        ll mid = (r + l) / 2;
  
        if (in_triangle(a, Point{0LL, 0LL}, b, hulls[i][mid])) {
          flag = true;
          break;
        }
  
        if (mid + 1 == (ll)hulls[i].size() || (!in_triangle(a, Point{0LL, 0LL}, b, hulls[i][mid + 1]) && sgn(hulls[i][mid + 1], a, hulls[i][mid]) == 1)) {
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }

      if (flag) break;
    }
  
    if (flag) {
      cout << "Y" << endl;
    } else {
      cout << "N" << endl;
    }
  }
  
  return 0;
}

Subtask #1

20.0 / 100.0

#	결과	실행 시간	메모리	Grader output
1	Correct	6 ms	348 KB	Output is correct
2	Correct	6 ms	512 KB	Output is correct
3	Incorrect	108 ms	2556 KB	Output isn't correct
4	Incorrect	374 ms	3320 KB	Output isn't correct
5	Incorrect	728 ms	6612 KB	Output isn't correct
6	Incorrect	777 ms	5320 KB	Output isn't correct
7	Incorrect	977 ms	6616 KB	Output isn't correct
8	Incorrect	474 ms	5600 KB	Output isn't correct
9	Incorrect	528 ms	5824 KB	Output isn't correct
10	Incorrect	585 ms	6656 KB	Output isn't correct