BZOJ 2178 Simpson积分

思路：

我发现能用Simpson积分水的题  好像都是裸题诶233333

//By SiriusRen
#include <bits/stdc++.h>
using namespace std;
#define pr pair<double,double>
const int N=1050;
int n,ban
;double inf=1e100,l=inf,r=-inf;pr p
;
struct Point{int x,y;};
double dis(Point a,Point b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
struct Circle{
Point p;int r;
pr f(double x){
if(r<=abs(p.x-x))return pr(0,0);
double t=sqrt(r*r-(p.x-x)*(p.x-x));
return pr(1.0*p.y-t,1.0*p.y+t);
}
}c
;
double Cut(double x){
double ret=0,last=-inf;int cnt=0;
for(int i=1;i<=n;i++){
p[++cnt]=c[i].f(x);
if(p[cnt]==pr(0,0))cnt--;
}
sort(p+1,p+1+cnt);
for(int i=1;i<=cnt;i++){
if(p[i].first>last)ret+=p[i].second-p[i].first,last=p[i].second;
else if(p[i].second>last)ret+=p[i].second-last,last=p[i].second;
}return ret;
}
double Simpson(double l,double r,double mid,double Cl,double Cr,double Cm){
double tCl=Cut((l+mid)/2),tCr=Cut((mid+r)/2);
double ans=(r-l)*(Cl+Cr+4*Cm)/6,lans=(mid-l)*(Cl+Cm+4*tCl)/6,rans=(r-mid)*(Cr+Cm+4*tCr)/6;
if(abs(lans+rans-ans)<1e-13)return ans;
else return Simpson(l,mid,(l+mid)/2,Cl,Cm,tCl)+Simpson(mid,r,(mid+r)/2,Cm,Cr,tCr);

}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d%d%d",&c[i].p.x,&c[i].p.y,&c[i].r);
l=min(l,(double)c[i].p.x-c[i].r);
r=max(r,(double)c[i].p.x+c[i].r);
}
for(int i=1;i<=n;i++){
if(ban[i])continue;
for(int j=i+1;j<=n;j++){
if(ban[j])continue;
if(dis(c[i].p,c[j].p)<=c[i].r-c[j].r)ban[j]=1;
}
}
for(int i=1;i<=n;i++)if(ban[i])swap(ban[i],ban
),swap(c[i--],c[n--]);
printf("%.3lf\n",Simpson(l,r,(l+r)/2,0,0,Cut((l+r)/2)));
}