K次圆覆盖问题

K次圆覆盖问题 模板

#include<bits/stdc++.h>
using namespace std;
const int maxn=1009;
const double eps=1e-8;
const double pi=acos(-1);
int dcmp(double x) {return fabs(x)<eps?0:(x<0?-1:1);}
struct circle {
	double x,y,r,angle;int d;
	circle() {}
	circle(double x,double y,double angle=0,int d=0)
		:x(x),y(y),angle(angle),d(d) {}
};
typedef circle Circle;
circle cir[maxn],tp[maxn<<1];
double area[maxn];
double sqr(double x) {return x*x;}
double dis(circle a,circle b) {return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
double cross(circle a,circle b,circle c){
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
int CirCrossCir(Circle p1, double r1,Circle p2, double r2,Circle &cp1,Circle &cp2) {
	double mx = p2.x - p1.x, sx = p2.x + p1.x, mx2 = mx * mx;
	double my = p2.y - p1.y, sy = p2.y + p1.y, my2 = my * my;
	double sq = mx2 + my2, d = -(sq - sqr(r1 - r2)) * (sq - sqr(r1 + r2));
	if (d + eps < 0)return 0;
	if (d < eps)d = 0;
	else d = sqrt(d);
	double x = mx * ((r1 + r2) * (r1 - r2) + mx * sx) + sx * my2;
	double y = my * ((r1 + r2) * (r1 - r2) + my * sy) + sy * mx2;
	double dx = mx * d, dy = my * d;
	sq *= 2;
	cp1.x = (x - dy) / sq;
	cp1.y = (y + dx) / sq;
	cp2.x = (x + dy) / sq;
	cp2.y = (y - dx) / sq;
	if (d > eps)return 2;
	else return 1;
}

bool circmp(const Circle& u, const Circle& v) {return dcmp(u.r - v.r) < 0;}
bool cmp(const Circle& u, const Circle& v) {
	if (dcmp(u.angle - v.angle))return u.angle < v.angle;
	return u.d > v.d;
}
double calc(Circle cir,Circle cp1,Circle cp2) {
	double ans = (cp2.angle - cp1.angle) * sqr(cir.r)
	             - cross(cir, cp1, cp2) + cross(Circle(0, 0), cp1, cp2);
	return ans / 2;
}

void CirUnion(Circle cir[], int n) {
	Circle cp1, cp2;
	sort(cir, cir + n, circmp);
	for (int i = 0; i < n; ++i)
		for (int j = i + 1; j < n; ++j)
			if (dcmp(dis(cir[i], cir[j]) + cir[i].r - cir[j].r) <= 0)
				cir[i].d++;
	for (int i = 0; i < n; ++i) {
		int tn = 0, cnt = 0;
		for (int j = 0; j < n; ++j) {
			if (i == j)continue;
			if (CirCrossCir(cir[i], cir[i].r, cir[j], cir[j].r,
			                cp2, cp1) < 2)
				continue;
			cp1.angle = atan2(cp1.y - cir[i].y, cp1.x - cir[i].x);
			cp2.angle = atan2(cp2.y - cir[i].y, cp2.x - cir[i].x);
			cp1.d = 1;
			tp[tn++] = cp1;
			cp2.d = -1;
			tp[tn++] = cp2;
			if (dcmp(cp1.angle - cp2.angle) > 0)
				cnt++;
		}
		tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, pi, -cnt);
		tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, -pi, cnt);
		sort(tp, tp + tn, cmp);
		int p, s = cir[i].d + tp[0].d;
		for (int j = 1; j < tn; ++j) {
			p = s;
			s += tp[j].d;
			area[p] += calc(cir[i], tp[j - 1], tp[j]);
		}
	}
}
int n;
void solve() {
	for(int i=0; i<n; i++) {
		scanf("%lf%lf%lf", &cir[i].x, &cir[i].y, &cir[i].r);
		cir[i].d = 1;
	}
	memset(area,0,sizeof area);
	CirUnion(cir,n);
	for(int i=1; i<=n; i++) {
		area[i]-=area[i+1];
		printf("[%d] = %.3f
", i, area[i]);  //n个圆i次覆盖的面积
	}
}
int main() {
	while(scanf("%d",&n)!=EOF)solve();
	return 0;
}