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[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(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[N]; 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[n]),swap(c[i--],c[n--]); printf("%.3lf\n",Simpson(l,r,(l+r)/2,0,0,Cut((l+r)/2))); }
时间: 2024-10-07 21:20:33