题面

题解

首先您得会用格林公式计算

这里只需要动态维护一下圆弧就可以了

时间复杂度\(O(n^2\log n)\)

//minamoto#include
    
     #define R register#define inline __inline__ __attribute__((always_inline))#define fp(i,a,b) for(R int i=(a),I=(b)+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)template
      
       inline bool cmax(T&a,const T&b){return a
       
        
         inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}using namespace std;const int N=2005;const double Pi=acos(-1.0);struct Point{    int x,y;    inline Point(){}    inline Point(R int xx,R int yy):x(xx),y(yy){}    inline Point operator +(const Point &b)const{return Point(x+b.x,y+b.y);}    inline Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}    inline double norm(){return sqrt(x*x+y*y);}    inline double ang(){return atan2(y,x);}};struct Cir{    Point p;int r;    inline double oint(R double t1,R double t2){        return r*(r*(t2-t1)+p.x*(sin(t2)-sin(t1))-p.y*(cos(t2)-cos(t1)));    }    inline bool in(const Cir &b)const{return (p-b.p).norm()+b.r<=r;}    inline bool out(const Cir &b)const{return r+b.r<=(p-b.p).norm();}}c[N];struct node{    double l,r;    inline node(R double ll,R double rr):l(ll),r(rr){}    inline bool operator <(const node &b)const{return r
         
          
           s[N];double res;int n;bool vis[N];typedef set
           
            ::iterator IT;inline void upd(R double &ang){ if(ang<-Pi)ang+=2*Pi; if(ang>Pi)ang-=2*Pi;}void remove(int id,double l,double r){ for(IT it=s[id].lower_bound(node(0,l)),tmp;it!=s[id].end()&&it->l
            
             l,nr=it->r; tmp=it,++tmp,s[id].erase(it),res-=c[id].oint(nl,nr); if(nl
             
              r)s[id].insert(node(r,nr)),res+=c[id].oint(r,nr); }}int main(){// freopen("testdata.in","r",stdin); scanf("%*d%d",&n); fp(i,1,n){ vis[i]=1; scanf("%d%d%d",&c[i].p.x,&c[i].p.y,&c[i].r); fp(j,1,i-1)if(vis[j]){ if(c[j].r>=c[i].r&&c[j].in(c[i])){vis[i]=0;break;} if(c[i].r>c[j].r&&c[i].in(c[j]))vis[j]=0,remove(j,-Pi,Pi); } if(!vis[i]){printf("%.10lf\n",res*0.5);continue;} res+=c[i].r*c[i].r*Pi*2,s[i].insert(node(-Pi,Pi)); fp(j,1,i-1)if(vis[j]&&!c[i].out(c[j])){ double dis=(c[i].p-c[j].p).norm(),ang,cur,l,r; cur=(c[j].p-c[i].p).ang(); ang=acos((dis*dis+c[i].r*c[i].r-c[j].r*c[j].r)/(2*c[i].r*dis)); l=cur-ang,r=cur+ang,upd(l),upd(r); if(l<=r)remove(i,l,r);else remove(i,-Pi,r),remove(i,l,Pi); cur=(c[i].p-c[j].p).ang(); ang=acos((dis*dis+c[j].r*c[j].r-c[i].r*c[i].r)/(2*c[j].r*dis)); l=cur-ang,r=cur+ang,upd(l),upd(r); if(l