【BZOJ】2178: 圆的面积并

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <sstream>
using namespace std;
typedef long long ll;
#define pb push_back
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline int getint() { static int r, k; r=0,k=1; static char c; c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }

const double eps=1e-6, PI=acos(-1);
int dcmp(double x) { return abs(x)<eps?0:(x<0?-1:1); }
double sqr(double x) { return x*x; }
struct iP { double x, y; iP(double _x=0, double _y=0) : x(_x), y(_y) {} };
double dis(iP &a, iP &b) { return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)); }
struct iC {
	iP p; double r;
	iP getP(double d) { return iP(p.x+cos(d)*r, p.y+sin(d)*r); }
	double areaH(double d) { return (d-sin(d))/2*sqr(r); }
};
double angle(iP &a, iP &b) {
	static double x, y;
	x=b.x-a.x; y=b.y-a.y;
	return atan2(y, x);
}
void CCi(iC &a, iC &b, double &ang1, double &ang2) {
	static double ang, d, R, r, da;
	d=dis(a.p, b.p); //dbg(d);
	ang=angle(a.p, b.p);
	R=a.r; r=b.r; //dbg((sqr(R)+sqr(d)-sqr(r))/2/R/d);
	da=acos((sqr(R)+sqr(d)-sqr(r))/2/R/d);
	ang1=ang-da;
	ang2=ang+da;
}

const int N=1105;
iC a[N];
int n, cnt;
bool nok[N];
struct dat { double pos; int k; }b[N*5];
bool cmp(const dat &a, const dat &b) { return a.pos<b.pos; }

int main() {
	read(n);
	for1(i, 1, n) scanf("%lf%lf%lf", &a[i].p.x, &a[i].p.y, &a[i].r);
	for1(i, 1, n) for1(j, 1, n) if(!nok[j] && i!=j) {
		double d=dis(a[i].p, a[j].p);
		if(dcmp(a[j].r-a[i].r-d)>=0) { nok[i]=1; break; }
	}
	double ang1, ang2, PI2=PI*2, ans=0;
	int sum=0;
	iP A, B;
	for1(i, 1, n) if(!nok[i]) {
		cnt=0;
		for1(j, 1, n) if(i!=j && !nok[j]) {
			if(dcmp(dis(a[i].p, a[j].p)-a[i].r-a[j].r)>=0) continue;
			CCi(a[i], a[j], ang1, ang2);
			if(dcmp(ang1)<0) ang1+=PI2;
			if(dcmp(ang2)<0) ang2+=PI2;
			if(dcmp(ang1-ang2)>0) {
				++cnt; b[cnt].pos=0; b[cnt].k=1;
				++cnt; b[cnt].pos=ang2; b[cnt].k=-1;
				++cnt; b[cnt].pos=ang1; b[cnt].k=1;
				++cnt; b[cnt].pos=PI2; b[cnt].k=-1;
			}
			else {
				++cnt; b[cnt].pos=ang1; b[cnt].k=1;
				++cnt; b[cnt].pos=ang2; b[cnt].k=-1;
			}
		}
		++cnt; b[cnt].pos=0; b[cnt].k=0;
		++cnt; b[cnt].pos=PI2; b[cnt].k=0;
		sort(b+1, b+1+cnt, cmp);
		sum=0;
		for1(j, 1, cnt-1) {
			sum+=b[j].k;
			if(!sum) {
				ans+=a[i].areaH(b[j+1].pos-b[j].pos);
				A=a[i].getP(b[j].pos);
				B=a[i].getP(b[j+1].pos);
				ans+=(A.x*B.y-A.y*B.x)/2;
			}
		}
	}
	printf("%.3f\n", ans);
	return 0;
}