SPOJ CIRU The area of the union of circles

You are given N circles and expected to calculate the area of the union of the circles !

Input

The first line is one integer n indicates the number of the circles. (1 <= n <= 1000)

Then follows n lines every line has three integers

Xi Yi Ri

indicates the coordinate of the center of the circle, and the radius. (|Xi|. |Yi|  <= 1000, Ri <= 1000)

Note that in this problem Ri may be 0 and it just means one point !

Output

The total area that these N circles with 3 digits after decimal point

Example

Input:
3

0 0 1
0 0 1
100 100 1

Output:
6.283

simpson自适应积分法

精度只需要1e-6，十分友好

调试语句懒得删

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 using namespace std;
 const double eps=1e-;
 const int INF=1e9;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*-''+ch;ch=getchar();}
     return x*f;
 }
 //
 struct cir{
     double x,y,r;
     friend bool operator < (const cir a,const cir b){return a.r<b.r;}
 }c[mxn];int cnt=;
 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));}
 //圆
 struct line{
     double l,r;
     friend bool operator <(const line a,const line b){return a.l<b.l;}
 }a[mxn],b[mxn];int lct=;
 double f(double x){
     int i,j;
     lct=;
     for(i=;i<=cnt;i++){//计算直线截得圆弧长度
         if(fabs(c[i].x-x)>=c[i].r)continue;
         double h= sqrt(c[i].r*c[i].r-(c[i].x-x)*(c[i].x-x));
         a[++lct].l=c[i].y-h;
         a[lct].r=c[i].y+h;
     }
     if(!lct)return ;
     double len=,last=-INF;
     sort(a+,a+lct+);
     for(i=;i<=lct;i++){//线段长度并
         if(a[i].l>last){len+=a[i].r-a[i].l;last=a[i].r;}
         else if(a[i].r>last){len+=a[i].r-last;last=a[i].r;}
     }
 //    printf("x:%.3f  len:%.3f\n",x,len);
     return len;
 }
 inline double sim(double l,double r){
     return (f(l)+*f((l+r)/)+f(r))*(r-l)/;
 }
 double solve(double l,double r,double S){
     double mid=(l+r)/;
     double ls=sim(l,mid);
     double rs=sim(mid,r);
     if(fabs(rs+ls-S)<eps)return ls+rs;
     return solve(l,mid,ls)+solve(mid,r,rs);
 }
 int n;
 double ans=;
 bool del[mxn];
 int main(){
     n=read();
     int i,j;
     double L=INF,R=-INF;
     for(i=;i<=n;i++){
         c[i].x=read();    c[i].y=read();    c[i].r=read();
 //        L=min(L,c[i].x-c[i].r);
 //        R=max(R,c[i].x+c[i].r);
     }
     //
     sort(c+,c+n+);
     for(i=;i<n;i++)
         for(j=i+;j<=n;j++){
 //            printf("%d %.3f %.3f %.3f %.3f\n",j,c[j].x,c[i].r,c[j].r,dist(c[i],c[j]));
             if(c[j].r-c[i].r>=dist(c[i],c[j]))
                 {del[i]=;break;}
         }
     for(i=;i<=n;i++)
         if(!del[i])c[++cnt]=c[i];
     //删去被包含的圆
 //    printf("cnt:%d\n",cnt);
     double tmp=-INF;int blct=;
     for(i=;i<=cnt;i++){
         b[++blct].l=c[i].x-c[i].r;
         b[blct].r=c[i].x+c[i].r;
     }
     sort(b+,b+blct+);
 //    printf("lct:%d\n",blct);
 //    int tlct=t;
     for(i=;i<=blct;i++){
 //        printf("%.3f %.3f\n",b[i].l,b[i].r);
 //        printf("tmp:%.3f\n",tmp);
         if(b[i].r<=tmp)continue;
         L=max(tmp,b[i].l);
 //        printf("%d: %.3f %.3f\n",i,L,a[i].r);
         ans+=solve(L,b[i].r,sim(L,b[i].r));
 //        printf("ANS:%.3f\n",ans);
 //        printf("nlct:%d\n",lct);
         tmp=b[i].r;
     }

 //    ans=solve(L,R,f((L+R)/2));
     printf("%.3f\n",ans);
     return ;
 }