ZOJ 3717 二分+2-sat判定。

好久没有2-sat了，此题当复习之用，二分求最大值+2-sat判断可行，此题主要跪于题意：The results should be rounded to three decimal
places. You should promise that there is still no overlap for any two balloons after rounded.
rounded是四舍五入的意思，按要求，3位之后要全部舍去，知道：printf（）,自动四舍五入。所有-0.0005

之后四舍五入即为要求。

#include<iostream>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
using namespace std;
int n;
struct points
{
    int x,y,z;
};
points point[402];
inline int getdis(points a,points b)   //三维
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
}
int vis[402];int dfn[402];int low[402];int scc[402];int numscc=0;
int times=0; stack<int>sta;int ins[402];
vector<vector<int> >v(402);
void clear()
{
    for(int i=0;i<2*n;i++)
    {
        ins[i]=vis[i]=dfn[i]=low[i]=scc[i]=0;
        v[i].clear();
    }
    numscc=times=0;
}
void tarjian(int u)
{
    dfn[u]=low[u]=++times;
    sta.push(u);
    ins[u]=1;
    int len=v[u].size();
    for(int i=0;i<len;i++)
      {
          int uu=v[u][i];
          if(!vis[uu])
          {
              vis[uu]=1;
              tarjian(uu);
             if(low[uu]<low[u])low[u]=low[uu];
          }
          else if(ins[uu]&&dfn[uu]<low[u])
             low[u]=dfn[uu];
      }
         if(low[u]==dfn[u])
        {
          int cur;
          numscc++;
          do
          {
              cur=sta.top();
              sta.pop();
              ins[cur]=0;
              scc[cur]=numscc;
          }while(cur!=u);
        }
}
bool check(double r)
{
    clear();
    for(int i=0;i<2*n-2;i++)
      for(int j=(i%2==0?i+2:i+1);j<2*n;j++)
    {
        double dis=0.00+getdis(point[i],point[j]);
        if(dis<4*r*r)
        {
            v[i].push_back(j^1);
            v[j].push_back(i^1);
        }
    }
    for(int i=0;i<2*n;i++)
      {
          if(!vis[i])
            {
                vis[i]=1;
                tarjian(i);
            }
      }
    for(int i=0;i<2*n;i+=2)
    {
        if(scc[i]==scc[i+1])
          return 0;
    }
    return 1;
}
int main()
{
    while(~scanf("%d",&n))
    {

        for(int i=0;i<2*n;i++)
          scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].z);
       double l=0,r=100000,mid;
       while(r-l>0.000001)      //二分。
       {
           mid=(r+l)/2;
           if(check(mid))
                  l=mid;
           else
              r=mid;
       }
       double ll=l-0.0005;
       printf("%.3lf
",ll);
    }
    return 0;
}