bzoj2178: 圆的面积并

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long double ld;
int n;
const ld pi=acos(-1.l),_2pi=pi*2;
struct itv{ld l,r;}is[2007];
bool operator<(itv x,itv y){return x.l<y.l;}
int ip;
ld ans=0;
ld maxs(ld&a,ld b){if(a<b)a=b;}
struct cir{
    int x,y,r;
    void init(){scanf("%d%d%d",&x,&y,&r);}
    bool in(cir w){
        int a=x-w.x,b=y-w.y;
        return sqrt(a*a+b*b)+r-1e-7l<w.r;
    }
    bool cross(cir w){
        int a=x-w.x,b=y-w.y;
        return sqrt(a*a+b*b)<r+w.r;
    }
    ld fix(ld x){
        while(x<0)x+=_2pi;
        while(x>_2pi)x-=_2pi;
        return x;
    }
    void cal(cir w){
        ld xd=w.x-x,yd=w.y-y,d=sqrt(xd*xd+yd*yd);
        ld a=atan2(yd,xd);
        ld b=acos((r*r+d*d-w.r*w.r)/(2*r*d));
        ld l=fix(a-b),r=fix(a+b);
        if(l<r)is[ip++]=(itv){l,r};
        else is[ip++]=(itv){0,r},is[ip++]=(itv){l,_2pi};
    }
    inline void g1(ld a){
        ans+=(a-sin(a))*r*r;
    }
    inline void g2(ld L,ld R){
        ans+=((x+cos(L)*r)*(y+sin(R)*r)-(x+cos(R)*r)*(y+sin(L)*r));
    }
    void get(){
        if(!ip){
            ans+=pi*r*r*2;
            return;
        }
        std::sort(is,is+ip);
        ld L,R=-1,R1;
        for(int i=0,j=0;i<ip;i=j){
            R1=R;L=is[i].l;R=is[i].r;
            while(j<ip&&is[j].l<=R)maxs(R,is[j++].r);
            if(R1!=-1)g1(L-R1),g2(R1,L);
        }
        g1(is[0].l+_2pi-R),g2(R,is[0].l+_2pi);
    }
}cs[1007];
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;++i)cs[i].init();
    for(int i=0;i<n;++i){
        for(int j=0;j<n;++j)if(i!=j&&cs[i].in(cs[j])){
            cs[i--]=cs[--n];
            break;
        }
    }
    for(int i=0;i<n;++i){
        ip=0;
        for(int j=0;j<n;++j)if(i!=j&&cs[i].cross(cs[j])){
            cs[i].cal(cs[j]);
        }
        cs[i].get();
    }
    printf("%.3Lf",ans/2.);
    return 0;
}