LOJ#2190. 「SHOI2014」信号增幅仪（最小圆覆盖）

太过爱你忘了你带给我的痛 2022-01-07 19:31 177阅读 0赞

## 题面 ##

[传送门][Link 1]

## 题解 ##

我连椭圆是个啥都不知道导致这么简单一道题我一点思路都没有……

我们把坐标系旋转一下，让半长轴成为新的\\(x\\)轴，也就是说所有点都绕原点逆时针旋转\\(360-a\\)度，然后再把所有点的\\(x\\)坐标变为原来的\\(\{1\\over p\}\\)，跑一个最小圆覆盖就行了

//minamoto #include<bits/stdc++.h> #define R register #define inline __inline__ __attribute__((always_inline)) #define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i) #define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i) #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v) using namespace std; char buf[1<<21],*p1=buf,*p2=buf; inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;} int read(){ R int res,f=1;R char ch; while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1); for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0'); return res*f; } double readdb() { R double x=0,y=0.1,f=1;R char ch; while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1); for(x=ch-'0';(ch=getc())>='0'&&ch<='9';x=x*10+ch-'0'); for(ch=='.'&&(ch=getc());ch>='0'&&ch<='9';x+=(ch-'0')*y,y*=0.1,ch=getc()); return x*f; } const int N=1e5+5;const double eps=1e-6,Pi=acos(-1.0); inline double reps(){return (1.0*rand()/RAND_MAX-0.5)*eps;} inline int sgn(R double x){return x<-eps?-1:x>eps;} struct node{ double x,y; inline node(){} inline node(R double xx,R double yy):x(xx),y(yy){} inline node operator +(const node &b)const{return node(x+b.x,y+b.y);} inline node operator -(const node &b)const{return node(x-b.x,y-b.y);} inline double operator *(const node &b)const{return x*b.y-y*b.x;} inline node operator *(const double &b)const{return node(x*b,y*b);} inline double operator ^(const node &b)const{return x*b.x+y*b.y;} inline double len2(){return x*x+y*y;} inline node rot(R double s,R double c){return node(x*c-y*s,x*s+y*c);} inline node rot90(){return node(-y,x);} inline void shake(){x+=reps(),y+=reps();} }p[N],o; struct Line{ node p,v; inline Line(){} inline Line(R node pp,R node vv):p(pp),v(vv){} friend node cross(const Line &a,const Line &b){return a.p+a.v*(b.v*(b.p-a.p)/(b.v*a.v));} }; node circle(const node &a,const node &b,const node &c){ return cross(Line((a+b)*0.5,(b-a).rot90()),Line((a+c)*0.5,(c-a).rot90())); } int n;double s,c,a,b; double calc(){ double r=0;o=node(0,0); random_shuffle(p+1,p+1+n); fp(i,1,n)if(sgn((p[i]-o).len2()-r)>0){ o=p[i],r=0; fp(j,1,i-1)if(sgn((p[j]-o).len2()-r)>0){ o=(p[i]+p[j])*0.5,r=(p[j]-o).len2(); fp(k,1,j-1)if(sgn((p[k]-o).len2()-r)>0) o=circle(p[i],p[j],p[k]),r=(p[k]-o).len2(); } } return sqrt(r); } int main(){ srand(20030719); // freopen("testdata.in","r",stdin); n=read(); fp(i,1,n)p[i].x=readdb(),p[i].y=readdb(),p[i].shake(); b=read(),a=read(),b=(360-b)/180.0*Pi,s=sin(b),c=cos(b),a=1.0/a; fp(i,1,n)p[i]=p[i].rot(s,c); fp(i,1,n)p[i].x=p[i].x*a; printf("%.3lf\n",calc()); return 0; }

转载于:https://www.cnblogs.com/bztMinamoto/p/10699737.html

[Link 1]: https://loj.ac/problem/2190