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-8;
const double inf = 1e5;
const double pi = acos(-1.0);
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    else if(x < 0)return -1;
    return 1;
}
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)) < 0 || sgn((p-e)*(s-e)) < 0)
            return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
};
const int MAXN = 2020;
Line line[MAXN];

double get(int i,double L)
{
    double l = 0, r = L;
    while(r - l >= eps)
    {
        double mid = (l + r)/2;
        double midmid = (r + mid)/2;
        if(line[i].dispointtoseg(Point(mid,0)) < line[i].dispointtoseg(Point(midmid,0)))
            r = midmid - eps;
        else l = mid + eps;
    }
    return l;
}
int n;
double L;

struct Node
{
    double a;
    int c;
    Node(double _a = 0,int _c = 0)
    {
        a = _a;
        c = _c;
    }
};
bool cmp(Node a,Node b)
{
    if(sgn(a.a - b.a) != 0)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)/2;
        if(line[i].dispointtoseg(Point(mid,0)) <= 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)/2;
        if(line[i].dispointtoseg(Point(mid,0)) <= R)
            l = mid + eps;
        else r = mid - eps;
    }
    return l;
}

bool gao(double r)
{
    vector<Node>vec;
    for(int i = 0;i < n;i++)
    {
        double tmp = get(i,L);
        if(sgn(line[i].dispointtoseg(Point(tmp,0)) - r) >= 0)
        {
            vec.push_back(Node(0,1));
            vec.push_back(Node(L,-1));
            continue;
        }
        if(sgn(line[i].dispointtoseg(Point(0,0)) - r) >= 0)
        {
            double tt = bin1(i,0,tmp,r);
            vec.push_back(Node(0,1));
            vec.push_back(Node(tt,-1));
        }
        if(sgn(line[i].dispointtoseg(Point(L,0)) - r) >= 0)
        {
            double tt = bin2(i,tmp,L,r);
            vec.push_back(Node(tt,1));
            vec.push_back(Node(L,-1));
        }
    }
    sort(vec.begin(),vec.end(),cmp);
    int cnt = 0;
    for(int i = 0;i < vec.size();i++)
    {
        cnt += vec[i].c;
        if(cnt >= n)return true;
    }
    return false;
}

double solve()
{
    double l = 0, r = inf;
    while(r-l >= eps)
    {
        double mid = (l+r)/2;
        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 = 0;i < n;i++)
            line[i].input();
        printf("%.3lf\n",solve());
    }
    return 0;
}