用read()会挂
二分半径，显然最优的是所有原都用这个最小半径，然后2-SAT把相交的圆建图，跑tarjan判一下可行性即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N=205;
int n,h[N],cnt,dfn[N],low[N],tot,s[N],top,bl[N],col;
bool v[N];
struct qwe
{int ne,to;
}e[N*N*8];
struct dian
{double x,y;
}a[N],b[N];
double dis(dian a,dian b)
{return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int read()
{int r=0,f=1;char p=getchar();while(p>='9'||p<'0'){if(p=='-')f=-1;p=getchar();}while(p>='0'&&p<='9'){r=r*10+p-48;p=getchar();}return r*f;
}
void add(int u,int v)
{//cerr<<u<<" "<<v<<endl;cnt++;e[cnt].ne=h[u];e[cnt].to=v;h[u]=cnt;
}
void tarjan(int u)
{dfn[u]=low[u]=++tot;v[s[++top]=u]=1;for(int i=h[u];i;i=e[i].ne){if(!dfn[e[i].to]){tarjan(e[i].to);low[u]=min(low[u],low[e[i].to]);}else if(v[e[i].to])low[u]=min(low[u],dfn[e[i].to]);}if(low[u]==dfn[u]){col++;while(s[top]!=u){bl[s[top]]=col;v[s[top--]]=0;}bl[s[top]]=col;v[s[top--]]=0;}
}
bool ok(double w)
{//cerr<<w<<endl;memset(h,0,sizeof(h));memset(v,0,sizeof(v));memset(dfn,0,sizeof(dfn));cnt=0,tot=0,top=0,col=0;for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++){if(dis(a[i],a[j])<2.0*w)add(i,j+n),add(j,i+n);if(dis(a[i],b[j])<2.0*w)add(i,j),add(j+n,i+n);if(dis(b[i],a[j])<2.0*w)add(i+n,j+n),add(j,i);if(dis(b[i],b[j])<2.0*w)add(i+n,j),add(j+n,i);}for(int i=1;i<=n+n;i++)if(!dfn[i])tarjan(i);for(int i=1;i<=n;i++)if(bl[i]==bl[i+n])return 0;return 1;
}
int main()
{while(~scanf("%d",&n)){for(int i=1;i<=n;i++)scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&b[i].x,&b[i].y);double l=0,r=1e4,ans=0;while(r-l>1e-3){double mid=(l+r)/2;if(ok(mid))l=mid,ans=mid;elser=mid;}printf("%.2f\n",ans);}return 0;
}