답안 #888166

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
888166	2023-12-16T09:18:37 Z	ad_red	Tri (CEOI09_tri)	C++17	20 / 100	1030 ms	6580 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 = 0, right_end = n - 1;
  
    // 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;
          break;
        }
      }
  
      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(right_end % sqrt_size == 0);
    // assert(left_start % sqrt_size == 0);
  
    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;
}

Compilation message

tri.cpp: In function 'int main()':
tri.cpp:197:104: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
  197 |         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) {
      |                                               ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Subtask #1

20.0 / 100.0

#	결과	실행 시간	메모리	Grader output
1	Correct	5 ms	344 KB	Output is correct
2	Correct	7 ms	348 KB	Output is correct
3	Incorrect	104 ms	2264 KB	Output isn't correct
4	Incorrect	393 ms	3520 KB	Output isn't correct
5	Incorrect	784 ms	6580 KB	Output isn't correct
6	Incorrect	874 ms	5380 KB	Output isn't correct
7	Incorrect	1030 ms	6440 KB	Output isn't correct
8	Incorrect	472 ms	5308 KB	Output isn't correct
9	Incorrect	525 ms	6200 KB	Output isn't correct
10	Incorrect	569 ms	6556 KB	Output isn't correct