Fruit Slicer Kattis - fruitslicer（计算几何+思维）

题意

给出n个圆的圆心坐标，每个圆的半径为1，问一条直线最多可穿过多少个圆

题解

类似于标准化的思想，首先以一个圆为基准，将其平移到圆心位置，同样的其他圆也要在同方向移动相同位移，之后再在其他的圆中任意找一个圆，以原点为圆心，将其旋转到与y=±1相切，为保持圆的相对位置不变，其他圆也要旋转相同角度，然后只需枚举每个圆的水平切线，再依据每条切线判断穿过圆的个数就方便多了，若一个圆的水平切线不满足-1≤y≤1，则舍弃，否则保存，最后对切线依次判断即可

#include <bits/stdc++.h>
#include <unordered_set>
#include <unordered_map>
#define _for(i, a, b) for(int i = a; i < b; ++i)
#define _rep(i, a, b) for(int i = a; i <= b; ++i)
#define closeIO ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug cout << "******************" << endl
#define FREE freopen("in.txt", "r", stdin)
#define FREO freopen("out.txt", "w", stdout)
#define ls l, m, rt << 1
#define rs m + 1, r, rt << 1 | 1
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef long double LD;
const int MAXN = 1e5 + 10;
const int MOD = 998244353;
const LD eps = 1e-10;
struct Point {
	LD x, y;
	Point(double x = 0, double y = 0) :x(x), y(y) {}
	void input() { scanf("%Lf%Lf", &x, &y); }
};
typedef Point Vector;
Vector operator+ (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator- (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator* (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator/ (Vector A, double p) { return Vector(A.x / p, A.y / p); }
bool operator< (const Point& a, const Point& b) { return a.x < b.x || a.x == b.x&&a.y < b.y; }
int dcmp(LD x) { if (fabs(x) < eps) return 0; return x < 0 ? -1 : 1; }
bool operator==(const Point& a, const Point& b) {
	return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
LD Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
LD Length(Vector A) { return sqrt(Dot(A, A)); }
LD Angle(Vector A, Vector B) {
	if (dcmp(Length(B)) == 0) return 0;
	return acos(Dot(A, B) / Length(A) / Length(B));
}
LD Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
LD Area2(Point A, Point B, Point C) { return Cross(B - A, C - A); }
Vector Rotate(Vector A, LD rad) {
	return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
}
Vector Normal(Vector A) {
	LD L = Length(A);
	return Vector(-A.y / L, A.x / L);
}
inline bool inside(LD x) {
	return dcmp(x + 1) >= 0 && dcmp(x - 1) <= 0;
}
int n;
Point P[105], Q[105];
int ans = 0;
int main() {
	scanf("%d", &n);
	_for(i, 0, n) P[i].input();
	if (n == 1) { printf("1\n"); return 0; }
	if (n == 2) { printf("2\n"); return 0; }
	sort(P, P + n);
	_for(i, 0, n) {
		Vector deltav(-P[i].x, -P[i].y);
		_for(j, 0, n) P[j] = P[j] + deltav;
		_for(j, i + 1, n) {
			LD rad = Angle(Vector(1, 0), Vector(P[j]));
			if (dcmp(Cross(Vector(1, 0), Vector(P[j]))) > 0) {
				_for(k, 0, n) Q[k] = Rotate(Vector(P[k]), -rad);
			}
			else {
				_for(k, 0, n) Q[k] = Rotate(Vector(P[k]), rad);
			}
			vector<LD> vec;
			int cnt;
			_for(k, 0, n) {
				if (k == i || k == j) continue;
				if (dcmp(fabs(Q[k].y) - 2) > 0) continue;
				if (inside(Q[k].y + 1)) vec.push_back(Q[k].y + 1);
				if (inside(Q[k].y - 1)) vec.push_back(Q[k].y - 1);
			}
			vec.push_back(1);
			vec.push_back(-1);
			_for(k, 0, vec.size()) {
				int cnt = 2;
				LD d = vec[k];
				_for(t, 0, n) {
					if (t == i || t == j) continue;
					if (dcmp(d + 1 - Q[t].y) >= 0 && dcmp(d - Q[t].y - 1) <= 0) ++cnt;
				}
				ans = max(ans, cnt);
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}