1 #include<iostream>  2 #include<cstdio>  3 #include<algorithm>  4 #include<cstring>  5 #include<cmath>  6 using namespace std;  7 const double eps=1e-6;  8 const int INF=1e9;  9 const int mxn=1010; 10 int read(){ 11     int x=0,f=1;char ch=getchar(); 12     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 13     while(ch>=‘0‘ && ch<=‘9‘){x=x*10-‘0‘+ch;ch=getchar();} 14     return x*f; 15 } 16 // 17 struct cir{ 18     double x,y,r; 19     friend bool operator < (const cir a,const cir b){return a.r<b.r;} 20 }c[mxn];int cnt=0; 21 inline double dist(cir a,cir b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));} 22 //圆 23 struct line{ 24     double l,r; 25     friend bool operator <(const line a,const line b){return a.l<b.l;} 26 }a[mxn],b[mxn];int lct=0; 27 double f(double x){ 28     int i,j; 29     lct=0; 30     for(i=1;i<=cnt;i++){//计算直线截得圆弧长度  31         if(fabs(c[i].x-x)>=c[i].r)continue; 32         double h= sqrt(c[i].r*c[i].r-(c[i].x-x)*(c[i].x-x)); 33         a[++lct].l=c[i].y-h; 34         a[lct].r=c[i].y+h; 35     } 36     if(!lct)return 0; 37     double len=0,last=-INF; 38     sort(a+1,a+lct+1); 39     for(i=1;i<=lct;i++){//线段长度并  40         if(a[i].l>last){len+=a[i].r-a[i].l;last=a[i].r;} 41         else if(a[i].r>last){len+=a[i].r-last;last=a[i].r;} 42     } 43 //    printf("x:%.3f  len:%.3f\n",x,len); 44     return len; 45 } 46 inline double sim(double l,double r){ 47     return (f(l)+4*f((l+r)/2)+f(r))*(r-l)/6; 48 } 49 double solve(double l,double r,double S){ 50     double mid=(l+r)/2; 51     double ls=sim(l,mid); 52     double rs=sim(mid,r); 53     if(fabs(rs+ls-S)<eps)return ls+rs; 54     return solve(l,mid,ls)+solve(mid,r,rs); 55 } 56 int n; 57 double ans=0; 58 bool del[mxn]; 59 int main(){ 60     n=read(); 61     int i,j; 62     double L=INF,R=-INF; 63     for(i=1;i<=n;i++){ 64         c[i].x=read();    c[i].y=read();    c[i].r=read(); 65 //        L=min(L,c[i].x-c[i].r); 66 //        R=max(R,c[i].x+c[i].r); 67     } 68     // 69     sort(c+1,c+n+1); 70     for(i=1;i<n;i++) 71         for(j=i+1;j<=n;j++){ 72 //            printf("%d %.3f %.3f %.3f %.3f\n",j,c[j].x,c[i].r,c[j].r,dist(c[i],c[j])); 73             if(c[j].r-c[i].r>=dist(c[i],c[j])) 74                 {del[i]=1;break;} 75         } 76     for(i=1;i<=n;i++) 77         if(!del[i])c[++cnt]=c[i]; 78     //删去被包含的圆 79 //    printf("cnt:%d\n",cnt); 80     double tmp=-INF;int blct=0; 81     for(i=1;i<=cnt;i++){ 82         b[++blct].l=c[i].x-c[i].r; 83         b[blct].r=c[i].x+c[i].r; 84     } 85     sort(b+1,b+blct+1); 86 //    printf("lct:%d\n",blct); 87 //    int tlct=t; 88     for(i=1;i<=blct;i++){ 89 //        printf("%.3f %.3f\n",b[i].l,b[i].r); 90 //        printf("tmp:%.3f\n",tmp); 91         if(b[i].r<=tmp)continue; 92         L=max(tmp,b[i].l); 93 //        printf("%d: %.3f %.3f\n",i,L,a[i].r); 94         ans+=solve(L,b[i].r,sim(L,b[i].r)); 95 //        printf("ANS:%.3f\n",ans); 96 //        printf("nlct:%d\n",lct); 97         tmp=b[i].r; 98     } 99     100     101 //    ans=solve(L,R,f((L+R)/2));102     printf("%.3f\n",ans);103     return 0;104 }