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=;
int n,ban[N];double inf=1e100,l=inf,r=-inf;pr p[N];
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(,);
        double t=sqrt(r*r-(p.x-x)*(p.x-x));
        return pr(1.0*p.y-t,1.0*p.y+t);
    }
}c[N];
double Cut(double x){
    double ret=,last=-inf;int cnt=;
    for(int i=;i<=n;i++){
        p[++cnt]=c[i].f(x);
        if(p[cnt]==pr(,))cnt--;
    }
    sort(p+,p++cnt);
    for(int i=;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)/),tCr=Cut((mid+r)/);
    double ans=(r-l)*(Cl+Cr+*Cm)/,lans=(mid-l)*(Cl+Cm+*tCl)/,rans=(r-mid)*(Cr+Cm+*tCr)/;
    if(abs(lans+rans-ans)<1e-)return ans;
    else return Simpson(l,mid,(l+mid)/,Cl,Cm,tCl)+Simpson(mid,r,(mid+r)/,Cm,Cr,tCr);

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