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-6;
const int INF=1e9;
const int mxn=1010;
int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0' && ch<='9'){x=x*10-'0'+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=0;
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=0;
double f(double x){
int i,j;
lct=0;
for(i=1;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 0;
double len=0,last=-INF;
sort(a+1,a+lct+1);
for(i=1;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)+4*f((l+r)/2)+f(r))*(r-l)/6;
}
double solve(double l,double r,double S){
double mid=(l+r)/2;
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=0;
bool del[mxn];
int main(){
n=read();
int i,j;
double L=INF,R=-INF;
for(i=1;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+1,c+n+1);
for(i=1;i<n;i++)
for(j=i+1;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]=1;break;}
}
for(i=1;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=0;
for(i=1;i<=cnt;i++){
b[++blct].l=c[i].x-c[i].r;
b[blct].r=c[i].x+c[i].r;
}
sort(b+1,b+blct+1);
//    printf("lct:%d\n",blct);
//    int tlct=t;
for(i=1;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 0;
}