hdu3622

题解：

2-sat

和上一题差不多

只不过是一个圆

卡精度卡了好久。。。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=405;
double x[N],y[N],aans;
int flag[N],n,ne[4*N*N],fi[N],zz[4*N*N],num;
int t,zhan[N],T,dfn[N],m,l,q,ans,low[N],an[N];
void jb(int x,int y)
{
    ne[++num]=fi[x];
    fi[x]=num;
    zz[num]=y;
}
void dfs(int x)
{
    low[x]=dfn[x]=++l;
    zhan[++t]=x;
    flag[x]=true;
    for (int i=fi[x];i!=0;i=ne[i])
     {
         if (an[zz[i]])continue;
        if(!dfn[zz[i]])dfs(zz[i]);
        if(!flag[zz[i]])low[x]=min(low[x],dfn[zz[i]]);else
        low[x]=min(low[x],low[zz[i]]);
     }
    if (dfn[x]==low[x])
     {
         ans++;
         while (zhan[t]!=x)
          {
              flag[zhan[t]]=false;
              an[zhan[t--]]=ans;
          }
         an[zhan[t--]]=ans;
         flag[x]=false;
     }
}
void init()
{
    ans=num=l=0;
    memset(fi,0,sizeof fi);
    memset(an,0,sizeof an);
    memset(dfn,0,sizeof dfn);
}
void build(double r)
{  
    init();  
    for (int i=0;i<2*n;i++)
     for (int j=i+1;j<2*n;j++)
      if (i%n!=j%n)
       if (sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]))<=r)
        {
            jb(i,(j+n)%(2*n));
            jb(j,(i+n)%(2*n));
        } 
}
int main()
{
    while (~scanf("%d",&n))
     {
        for (int i=0;i<n;i++)
         scanf("%lf%lf%lf%lf",&x[i],&y[i],&x[i+n],&y[i+n]);
        double l=0,r=2e9;aans=0;
        for (int kk=0;kk<50;kk++)
         {
             double mid=(l+r)/2,k=0;
             build(mid);
             for (int i=0;i<2*n;i++)
              if (!dfn[i])dfs(i);
             for (int i=0;i<n;i++)
             if (an[i]==an[i+n])
              {r=mid;k=1;break;}
            if (!k)l=aans=mid;  
         }
        printf("%.2lf
",aans/2); 
     }
}