【BZOJ】【2178】圆的面积并

自适应辛普森积分

---

　　Orz Hzwer

　　辛普森真是个强大的东西……很多东西都能积= =

　　这题的正解看上去很鬼畜，至少我这种不会计算几何的渣渣是写不出来……（对圆的交点求图包，ans=凸包的面积+一堆弓形的面积，另外还有中空的情况……那种凸包怎么求啊喂！）

 /**************************************************************
     Problem: 2178
     User: Tunix
     Language: C++
     Result: Accepted
     Time:8808 ms
     Memory:1372 kb
 ****************************************************************/

 //BZOJ 2178
 #include<cmath>
 #include<vector>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 #define pb push_back
 using namespace std;
 inline int getint(){
     int v=,sign=; char ch=getchar();
     while(ch<''||ch>''){ if (ch=='-') sign=-; ch=getchar();}
     while(ch>=''&&ch<=''){ v=v*+ch-''; ch=getchar();}
     return v*sign;
 }
 const int N=,INF=~0u>>;
 const double eps=1e-;
 typedef long long LL;
 typedef double db;
 /******************tamplate*********************/
 int n,top,st,ed;
 db xl[N],xr[N],ans;
 bool del[N];
 struct data{db x,y,r;}t[N],sk[N];
 struct line{db l,r;}p[N];
 db dis(data a,data b){
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 bool cmp1(data a,data b){return a.r<b.r;}
 bool cmp2(data a,data b){return a.x-a.r<b.x-b.r;}
 bool cmp3(line a,line b){return a.l<b.l;}
 void init(){
     scanf("%d",&n);
     F(i,,n) scanf("%lf%lf%lf",&t[i].x,&t[i].y,&t[i].r);
     sort(t+,t+n+,cmp1);
     F(i,,n) F(j,i+,n)
         if(dis(t[i],t[j])<=t[j].r-t[i].r){
             del[i]=; break;
         }
     F(i,,n)if(!del[i]) sk[++top]=t[i];n=top;
     sort(sk+,sk+n+,cmp2);
 }
 db getf(db x){
     int j,sz=; db r,len=,dis;
     F(i,st,ed){
         if (x<=xl[i]||x>=xr[i])continue;
         dis=sqrt(sk[i].r-(x-sk[i].x)*(x-sk[i].x));
         p[++sz].l=sk[i].y-dis; p[sz].r=sk[i].y+dis;
     }
     sort(p+,p+sz+,cmp3);
     F(i,,sz){
         r=p[i].r;
         for(j=i+;j<=sz;j++){
             if(p[j].l>r)break;
             if (r<p[j].r) r=p[j].r;
         }
         len+=r-p[i].l; i=j-;
     }
     return len;
 }
 db cal(db l,db fl,db fmid,db fr){
     return (fl+fmid*+fr)*l/;
 }
 db simpson(db l,db mid,db r,db fl,db fmid,db fr,db s){
     db m1=(l+mid)/,m2=(mid+r)/;
     db f1=getf(m1),f2=getf(m2);
     db g1=cal(mid-l,fl,f1,fmid),g2=cal(r-mid,fmid,f2,fr);
     if (fabs(g1+g2-s)<eps)return g1+g2;
     return simpson(l,m1,mid,fl,f1,fmid,g1)+
            simpson(mid,m2,r,fmid,f2,fr,g2);
 }
 void work(){
     int i,j;
     db l,r,mid,fl,fr,fmid;
     F(i,,n){
         xl[i]=sk[i].x-sk[i].r;
         xr[i]=sk[i].x+sk[i].r;
         sk[i].r*=sk[i].r;
     }
     F(i,,n){
         l=xl[i]; r=xr[i];
         for(j=i+;j<=n;j++){
             if (xl[j]>r) break;
             if (xr[j]>r) r=xr[j];
         }
         st=i; ed=j-;i=j-;
         mid=(l+r)/;
         fl=getf(l); fr=getf(r); fmid=getf(mid);
         ans+=simpson(l,mid,r,fl,fmid,fr,cal(r-l,fl,fmid,fr));
     }
 }
 int main(){
 #ifndef ONLINE_JUDGE
     freopen("2178.in","r",stdin);
     freopen("2178.out","w",stdout);
 #endif
     init();
     work();
     printf("%.3lf",ans);
     return ;
 }

2178: 圆的面积并

Time Limit: 20 Sec  Memory Limit: 259 MB
Submit: 908  Solved: 218
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Description

给出N个圆，求其面积并

Input

先给一个数字N ,N< = 1000
接下来是N行是圆的圆心，半径，其绝对值均为小于1000的整数

Output

面积并，保留三位小数

Sample Input

721

。。。。。。

Sample Output

12707279.690

HINT

Source

鸣谢 Aekdycoin

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