답안 #888609

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
888609	2023-12-18T01:41:13 Z	ad_red	Tri (CEOI09_tri)	C++17	100 / 100	1055 ms	6784 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(hull[(ll)hull.size() - 2], p, hull.back()) <= 0LL) {
        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, right_end;
  
    // 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;
        } else {
          r = mid;
        }
      }

      right_end = l;
    }
  
    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

100.0 / 100.0

#	결과	실행 시간	메모리	Grader output
1	Correct	6 ms	348 KB	Output is correct
2	Correct	6 ms	348 KB	Output is correct
3	Correct	93 ms	2308 KB	Output is correct
4	Correct	503 ms	3528 KB	Output is correct
5	Correct	1055 ms	6784 KB	Output is correct
6	Correct	812 ms	5360 KB	Output is correct
7	Correct	1010 ms	6648 KB	Output is correct
8	Correct	467 ms	5596 KB	Output is correct
9	Correct	537 ms	6112 KB	Output is correct
10	Correct	625 ms	6428 KB	Output is correct