D. Capture Stars

D. Capture Stars

反演

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定义

若 \(OP\cdot OP' = r^2\) , 则称 \(P\) 与 \(P'\) 关于 \(O\) 互为反演，反演半径为 \(r\)

该题反演后，转化为很简单的问题

#include<bits/stdc++.h>
using namespace std;

int T, n, R, r, cnt;
const int N = 1e4 + 10;
double x[N], y[N];
const double eps = 1e-8;
struct node {
	double y;
	int val;
	bool operator < (const node& b)const {
		return fabs(y - b.y) < eps ? val > b.val : y < b.y;
	}
}A[N<<1];
void insert(int idx) {
	double d = fabs(1.0 * R * R / r + R - x[idx]);
	double dy = sqrt((1.0 * R * R / r - R) * (1.0 * R * R / r - R) - d * d);
	A[cnt++] = { y[idx] + dy ,-1 };
	A[cnt++] = { y[idx] - dy,1 };
}
int main() {
	scanf("%d", &T);
	while (T--) {
		cnt = 0;
		scanf("%d%d%d", &n, &R, &r);
		for (int i = 0; i < n; i++) {
			scanf("%lf%lf", x + i, y + i);
			double dis = sqrt(x[i] * x[i] + y[i] * y[i]);
			double t = 4 * R * R / (dis*dis);
			x[i] *= t; y[i] *= t;
			insert(i);
		}
		//x = R*R/r + R
		int ans = 0, now = 0;
		sort(A, A + cnt);
		for (int i = 0; i < cnt; i++) {
			now += A[i].val;
			ans = max(ans, now);
		}
		printf("%d\n", ans);
	}
}