UVALive 4818 - Largest Empty Circle on a Segment （计算几何）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2819

题目意思：

给了平面上的n条线段：

让你在x轴上[0,L]的范围内找一个点作为圆心，画一个圆，这个圆不能把线段包含在里面去。

求最大的半径。

求解：

二分最终的答案，求解。

对于二分的半径值，对每条线段找出对于x位置上满足半径要求的段。

看n条线段有没有交集就可以了。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2014/5/10 23:18:09
 File Name     :E:\2014ACM\区域赛练习\2010\2010SouthEastern_European_Region\C.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 const double eps = 1e-;
 const double inf = 1e5;
 const double pi = acos(-1.0);
 int sgn(double x)
 {
     if(fabs(x) < eps)return ;
     else if(x < )return -;
     return ;
 }
 struct Point
 {
     double x,y;
     Point(){}
     Point(double _x,double _y)
     {
         x = _x;
         y = _y;
     }
     void input()
     {
         scanf("%lf%lf",&x,&y);
     }
     Point operator -(const Point &b)const
     {
         return Point(x-b.x,y-b.y);
     }
     double operator *(const Point &b)const
     {
         return x*b.x + y*b.y;
     }
     double operator ^(const Point &b)const
     {
         return x*b.y - y*b.x;
     }
     double len()
     {
         return hypot(x,y);
     }
     double distance(Point p)
     {
         return hypot(x-p.x,y-p.y);
     }
 };
 struct Line
 {
     Point s,e;
     Line(){}
     Line(Point _s,Point _e)
     {
         s = _s;
         e = _e;
     }
     void input()
     {
         s.input();
         e.input();
     }
     double length()
     {
         return s.distance(e);
     }
     double dispointtoline(Point p)
     {
         return fabs((p-s)^(e-s))/length();
     }
     double dispointtoseg(Point p)
     {
         if(sgn((p-s)*(e-s)) <  || sgn((p-e)*(s-e)) < )
             return min(p.distance(s),p.distance(e));
         return dispointtoline(p);
     }
 };
 const int MAXN = ;
 Line line[MAXN];

 double get(int i,double L)
 {
     double l = , r = L;
     while(r - l >= eps)
     {
         double mid = (l + r)/;
         double midmid = (r + mid)/;
         if(line[i].dispointtoseg(Point(mid,)) < line[i].dispointtoseg(Point(midmid,)))
             r = midmid - eps;
         else l = mid + eps;
     }
     return l;
 }
 int n;
 double L;

 struct Node
 {
     double a;
     int c;
     Node(double _a = ,int _c = )
     {
         a = _a;
         c = _c;
     }
 };
 bool cmp(Node a,Node b)
 {
     if(sgn(a.a - b.a) != )return a.a < b.a;
     else return a.c > b.c;
 }

 double bin1(int i,double l,double r,double R)
 {
     while(r-l >= eps)
     {
         double mid = (l+r)/;
         if(line[i].dispointtoseg(Point(mid,)) <= R)
             r = mid - eps;
         else l = mid + eps;
     }
     return l;
 }
 double bin2(int i,double l,double r,double R)
 {
     while(r-l >= eps)
     {
         double mid = (l+r)/;
         if(line[i].dispointtoseg(Point(mid,)) <= R)
             l = mid + eps;
         else r = mid - eps;
     }
     return l;
 }

 bool gao(double r)
 {
     vector<Node>vec;
     for(int i = ;i < n;i++)
     {
         double tmp = get(i,L);
         if(sgn(line[i].dispointtoseg(Point(tmp,)) - r) >= )
         {
             vec.push_back(Node(,));
             vec.push_back(Node(L,-));
             continue;
         }
         if(sgn(line[i].dispointtoseg(Point(,)) - r) >= )
         {
             double tt = bin1(i,,tmp,r);
             vec.push_back(Node(,));
             vec.push_back(Node(tt,-));
         }
         if(sgn(line[i].dispointtoseg(Point(L,)) - r) >= )
         {
             double tt = bin2(i,tmp,L,r);
             vec.push_back(Node(tt,));
             vec.push_back(Node(L,-));
         }
     }
     sort(vec.begin(),vec.end(),cmp);
     int cnt = ;
     for(int i = ;i < vec.size();i++)
     {
         cnt += vec[i].c;
         if(cnt >= n)return true;
     }
     return false;
 }

 double solve()
 {
     double l = , r = inf;
     while(r-l >= eps)
     {
         double mid = (l+r)/;
         if(gao(mid))l = mid+eps;
         else r = mid - eps;
     }
     return l;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%lf",&n,&L);
         for(int i = ;i < n;i++)
             line[i].input();
         printf("%.3lf\n",solve());
     }
     return ;
 }