답안 #888171

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
888171	2023-12-16T09:29:04 Z	ad_red	Tri (CEOI09_tri)	C++17	20 / 100	1006 ms	7008 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 = 0;
      ll r = n - 1;
    
      while (l <= r) {
        ll mid = (l + r) / 2;
        if (sgn(a, Point{0LL, 0LL}, points[mid]) >= 0) {
          l = mid + 1;
        } else {
          r = mid - 1;
          left_start = mid;
        }
      }
    }
  
    {
      ll l = 0;
      ll r = n - 1;
    
      while (l <= r) {
        ll mid = (l + r) / 2;
        if (sgn(b, Point{0LL, 0LL}, points[mid]) >= 0) {
          l = mid + 1;
          right_end = mid;
        } else {
          r = mid - 1;
        }
      }
    }
  
    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();
  
      while (r - l > 1) {
        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;
        } else {
          l = mid;
        }
      }

      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	344 KB	Output is correct
2	Correct	6 ms	348 KB	Output is correct
3	Incorrect	95 ms	2300 KB	Output isn't correct
4	Incorrect	375 ms	3272 KB	Output isn't correct
5	Incorrect	723 ms	7008 KB	Output isn't correct
6	Incorrect	811 ms	5364 KB	Output isn't correct
7	Incorrect	1006 ms	6920 KB	Output isn't correct
8	Incorrect	469 ms	5336 KB	Output isn't correct
9	Incorrect	531 ms	6228 KB	Output isn't correct
10	Incorrect	586 ms	6448 KB	Output isn't correct