| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 887836 | ad_red | Tri (CEOI09_tri) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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;
ll 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 (l >= r) {
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;
}