Problem B. Geometry Problem
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88926#problem/B

Description

Peter is studying in the third grade of elementary school. His teacher of geometry often gives him difficult home tasks. At the last lesson the students were studying circles. They learned how to draw circles with compasses. Peter has completed most of his homework and now he needs to solve the following problem. He is given two segments. He needs to draw a circle which intersects interior of each segment exactly once. The circle must intersect the interior of each segment, just touching or passing through the end of the segment is not satisfactory. Help Peter to complete his homework.

Input

The input file contains several test cases. Each test case consists of two lines. The first line of the test case contains four integer numbers x11, y11, x12, y12 — the coordinates of the ends of the first segment. The second line contains x21, y21, x22, y22 and describes the second segment in the same way. Input is followed by two lines each of which contains four zeroes these lines must not be processed. All coordinates do not exceed 102 by absolute value.

Output

For each test case output three real numbers — the coordinates of the center and the radius of the circle. All numbers in the output file must not exceed 1010 by their absolute values. The jury makes all comparisons of real numbers with the precision of 10−4 .

Sample Input

0 0 0 4
1 0 1 4
0 0 0 0
0 0 0 0

Sample Output

0.5 0 2

HINT

题意

给你两个线段,让你构造一个圆,与每个线段都只相交一次

题解

首先如何判断这个线段和圆相交了一次:一个端点在圆内,一个在圆外

然后我们枚举四个点的中点,半径就取中点到端点的最小值,然后再随便加上一个0.005就好了

就AC了……

加0.05会WA8

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200051
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** struct node
{
double x,y;
};
node kiss1[];
node kiss2[];
double dis(node a,node b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int check1(double r,double x,double y)
{
int flag=;
int flag1=;
node ttt;
ttt.x=x,ttt.y=y;
for(int i=;i<;i++)
{
if(dis(ttt,kiss1[i])<r)
flag++;
if(dis(ttt,kiss2[i])<r)
flag1++;
}
if(flag==&&flag1==)
return ;
return ;
}
int check(double x,double y)
{
double ans=inf;
node a;
a.x=x,a.y=y;
for(int i=;i<;i++)
{
ans = min(ans,dis(a,kiss1[i]));
ans = min(ans,dis(a,kiss2[i]));
}
ans+=0.005;
if(check1(ans,x,y))
{
printf("%.10f %.10f %.10f\n",x,y,ans);
return ;
}
return ;
}
int main()
{
srand((unsigned)time(NULL));
freopen("geometry.in","r",stdin);
freopen("geometry.out","w",stdout);
while()
{
for(int i=;i<;i++)
cin>>kiss1[i].x>>kiss1[i].y;
for(int i=;i<;i++)
cin>>kiss2[i].x>>kiss2[i].y;
if(kiss1[].x==&&kiss1[].y==&&kiss2[].x==&&kiss2[].y==&&kiss1[].x==&&kiss1[].y==&&kiss2[].x==&&kiss2[].y==)
break;
node a,b;
double ans=inf;
int flag = ;
for(int i=;i<;i++)
{
for(int j=;j<;j++)
{
a = kiss1[i],b = kiss2[j];
double ansx=(a.x+b.x)/2.0;
double ansy=(a.y+b.y)/2.0;
if(check(ansx,ansy)==)
{
flag = ;
break;
}
}
if(!flag)
break;
}
if(!flag)
continue;
}
}

Codeforces Gym 100338B Geometry Problem 计算几何的更多相关文章

  1. Codeforces Gym 100338B Spam Filter 字符串哈希+贝叶斯公式

    原题链接:http://codeforces.com/gym/100338/attachments/download/2136/20062007-winter-petrozavodsk-camp-an ...

  2. Hdu-6242 2017CCPC-哈尔滨站 M.Geometry Problem 计算几何 随机

    题面 题意:给你n个点,让你找到一个圆,输出圆心,和半径,使得有超过一半的点刚好在圆上.n<=1e5,题目保证了有解 题解:刚开始看着很不可做的样子,但是多想想,三点确定一个圆,三点啊! 现在有 ...

  3. Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 离散化 排列组合

    E. Mike and Geometry Problem 题目连接: http://www.codeforces.com/contest/689/problem/E Description Mike ...

  4. hdu 1086:You can Solve a Geometry Problem too(计算几何,判断两线段相交,水题)

    You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/3 ...

  5. Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 【逆元求组合数 && 离散化】

    任意门:http://codeforces.com/contest/689/problem/E E. Mike and Geometry Problem time limit per test 3 s ...

  6. codeforces 361 E - Mike and Geometry Problem

    原题: Description Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him ...

  7. CodeForces 689E Mike and Geometry Problem (离散化+组合数)

    Mike and Geometry Problem 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/I Description M ...

  8. Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 离散化+逆元

    E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input ...

  9. HDU1086 You can Solve a Geometry Problem too(计算几何)

    You can Solve a Geometry Problem too                                         Time Limit: 2000/1000 M ...

随机推荐

  1. 【转】第一次使用Android Studio时你应该知道的一切配置(三):gradle项目构建

    原文网址:http://www.cnblogs.com/smyhvae/p/4456420.html [声明] 欢迎转载,但请保留文章原始出处→_→ 生命壹号:http://www.cnblogs.c ...

  2. css中将div定位居中

    一直,我是认为定一个width,然后写一句margin:0 auto,就可以,但是有时也会不管用. 例如当我要定一个宽度为700的div,用相对定位定在中间.任你怎么拉伸都是居中.而position: ...

  3. android学习笔记五

    Android中的category大全 Api Level 3(SDK 1.5)和Api Level 4(SDK 1.6): android.intent.category.ALTERNATIVE a ...

  4. LXD 2.0 系列(二):安装与配置

    导读 简单来说,LXD是一个守护进程,为LXC容器的管理提供一组REST API.主要目标是提供一种类虚拟机的用户体验,是一种第三方的容器管理工具.下面呢,我们来介绍LXD 2.0 的安装与配置 安装 ...

  5. Eclipse “Invalid Project Description” when creating new project from existing source

    1) File>Import>General>Existing Project into Workspace2) File>Import>Android>Exist ...

  6. (转载) 数组a[]={3,5,2,4,1,8},要求从a中找出所有“和”等于10的子集

    背包问题.     不过就这道题目本身而言,由于集合a中只要6个元素,而不是成千上万,所以可以使用更直观的办法:     只要你能通过程序给出数组a中元素所组成的集合的所有的子集合(幂集),那么只需在 ...

  7. [持续更新]android stduio的一些小技巧

    1.导入第三方jar 无法和eclipse一样可以直接刷新就可以导入libs里的jar,要到jar文件上右键按add jar. 2.其他依赖项目里有support-v4包的时候 在Libs里还是要有s ...

  8. HDU 1892-See you(二维BIT)

    题意: 最多1000*1000的方格,各方格开始有一本书 有四种操作:对指定方格把书拿走或向里面放书,从一个方格那一定量的书放到另一个方格,查询给定对角线顶点的坐标的矩形范围内包含的书的总数 分析: ...

  9. java易混淆概念之类变量、成员变量、局部变量

      类变量.成员变量.局部变量 类变量(也叫静态变量)是类中独立于方法之外的变量,用static 修饰.(static表示“全局的”.“静态的”,用来修饰成员变量和成员方法,或静态代码块(静态代码块独 ...

  10. 重新执笔,已是大三!Jekyll自定义主题开发

    前言 “一转眼忘了时间 丢了感觉 黑了世界 再逞强 再疯狂 也会伤 不知 不觉 后知 后觉 然后 发现 失去 知觉 ”——<一吻不天荒> 感言 时间是把双刃剑,什么解决不了,忧烦的,慢慢变 ...