HDU 3264 Open-air shopping malls

题目链接

题意：给定一些圆，求以一个圆的圆心为圆心，自己定一个半径，使得和其他所有圆交面积都大于该圆的一半，求这个半径的最小值

思路：很显然的二分半径，判断方法就枚举一个圆心，然后和每个圆求圆交面积即可

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 25;
const double pi = acos(-1.0);
const double eps = 1e-8;
int t, n;
struct Circle {
	double x, y, r;
	void read() {
		scanf("%lf%lf%lf", &x, &y, &r);
	}
} c[N], o;

double dis(double x1, double y1, double x2, double y2) {
	double dx = x1 - x2;
	double dy = y1 - y2;
	return sqrt(dx * dx + dy * dy);
}

double area(Circle c) {
	return c.r * c.r * pi;
}

double dcmp(double x) {
	if (fabs(x) < eps) return 0.0;
	return x;
}

double cal(Circle c1, Circle c2) {
	double a = dis(c1.x, c1.y, c2.x, c2.y), b = c1.r, c = c2.r;
	double r1 = c1.r, r2 = c2.r;
	if (dcmp(a - b - c) >= 0) return 0.0;
	double minr = min(b, c);
	if (dcmp(a - fabs(b - c)) <= 0) return minr * minr * pi;
	double cta1 = acos((a * a + b * b - c * c) / 2 / (a * b));
	double cta2 = acos((a * a + c * c - b * b) / 2 / (a * c));
	double s1 = r1 * r1 * cta1 - r1 * r1 * sin(cta1) * (a * a + b * b - c * c) / 2 / (a * b);
	double s2 = r2 * r2 * cta2 - r2 * r2 * sin(cta2) * (a * a + c * c - b * b) / 2 / (a * c);
	return s1 + s2;
}

bool check() {
	for (int i = 0; i < n; i++)
		if (cal(o, c[i]) < area(c[i]) / 2) return false;
	return true;
}

bool judge(double r) {
	for (int i = 0; i < n; i++) {
		o.x = c[i].x; o.y = c[i].y; o.r = r;
		if (check()) return true;
	}
	return false;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
			c[i].read();
		double l = 0, r = 1e5;
		for (int i = 0; i < 100; i++) {
			double mid = (l + r) / 2;
			if (judge(mid)) r = mid;
			else l = mid;
		}
		printf("%.4lf\n", l);
	}
	return 0;
}