Problem Description
Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with coordinates (x1,y1) , (x2,y2) , and (x3,y3) . Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Could you tell him if it is outside the circle or not?
 
Input
The first line contains an integer T , indicating that there are T(T≤30) cases.
In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y

, as described above.
The absolute values of integers in input are less than or equal to 1,000,000,000,000

.
It is guaranteed that, any three of the four positions do not lie on a straight line.

 
Output
For each case, output "Accepted" if the position is outside the circle, or "Rejected" if the position is on or inside the circle.
 
Sample Input
3
-2 0 0 -2 2 0 2 -2
-2 0 0 -2 2 0 0 2
-2 0 0 -2 2 0 1 1
 
Sample Output
Accepted
Rejected
Rejected
 
Source
【题意】:给出四个点,问你第四个点是否在前三个点构成的圆内,若在圆外输出"Accepted",否则输出"Rejected",题目保证前三个点不在一条直线上。
【分析】:就是求出三个点外接圆的圆心和半径判断下。精度问题需要用Java大数。已知三点坐标,求外接圆圆心坐标与半径。三点构圆的圆心和半径是能够推导出公式的圆弧方向判断方法和三点确定一个圆的计算方法 高精度问题也只用BigInteger解决即可。
【彩蛋】:

实际上有更简便的方法,直接能用更直接的公式算出圆心 (x0, y0) 和半径的平方 r^2

x0 = ((y2-y1)*(y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1))/(2.0*((x3-x1)*(y2-y1)-(x2-x1)*(y3-y1)));

y0 = ((x2-x1)*(x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1))/(2.0*((y3-y1)*(x2-x1)-(y2-y1)*(x3-x1)));

r^2= (x1-x0)*(x1-x0)+(y1-y0)*(y1-y0);

里面涉及除法,那就用BigDecimal就能解决了,参考http://blog.csdn.net/cillyb/article/details/78012069

 
import java.math.*;
import java.util.*;
import java.io.*; public class Main{
public static void main(String[] args){
Scanner cin=new Scanner(System.in);
int t=cin.nextInt();
while(t-->)
{
BigDecimal px1, px2, px3, py1, py2, py3, px, py;
px1=cin.nextBigDecimal();
py1=cin.nextBigDecimal();
px2=cin.nextBigDecimal();
py2=cin.nextBigDecimal();
px3=cin.nextBigDecimal();
py3=cin.nextBigDecimal();
px=cin.nextBigDecimal();
py=cin.nextBigDecimal();
BigDecimal a, b, c, d, e, f, px0, py0, r,dis;
a=px1.subtract(px2);
b=py1.subtract(py2);
c=px1.subtract(px3);
d=py1.subtract(py3);
e=px1.multiply(px1).subtract(px2.multiply(px2)).multiply(BigDecimal.valueOf(0.5)).subtract(py2.multiply(py2).subtract(py1.multiply(py1)).multiply(BigDecimal.valueOf(0.5)));
f=px1.multiply(px1).subtract(px3.multiply(px3)).multiply(BigDecimal.valueOf(0.5)).subtract(py3.multiply(py3).subtract(py1.multiply(py1)).multiply(BigDecimal.valueOf(0.5)));
px0=b.multiply(f).subtract(d.multiply(e)).divide(b.multiply(c).subtract(a.multiply(d)),,BigDecimal.ROUND_HALF_UP);
py0=c.multiply(e).subtract(a.multiply(f)).divide(b.multiply(c).subtract(a.multiply(d)),,BigDecimal.ROUND_HALF_UP);
r=px1.subtract(px0).multiply(px1.subtract(px0)).add(py1.subtract(py0).multiply(py1.subtract(py0)));
dis=px.subtract(px0).multiply(px.subtract(px0)).add(py.subtract(py0).multiply(py.subtract(py0)));
if(dis.compareTo(r)==)
System.out.println("Accepted");
else
System.out.println("Rejected"); }
}
}

JAVA高精度

import java.math.BigDecimal;
import java.util.Scanner; public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
BigDecimal x1, y1, x2, y2, x3, y3, x4, y4;
int _;
_ = sc.nextInt();
while(_-- != )
{
x1 = sc.nextBigDecimal();
y1 = sc.nextBigDecimal();
x2 = sc.nextBigDecimal();
y2 = sc.nextBigDecimal();
x3 = sc.nextBigDecimal();
y3 = sc.nextBigDecimal();
x4 = sc.nextBigDecimal();
y4 = sc.nextBigDecimal();
BigDecimal t;
if(y3.equals(y1))
{
t = y2;
y2 = y3;
y3 = t; t = x2;
x2 = x3;
x3 = t;
}
BigDecimal t1 = (y3.subtract(y1)).multiply(y2.multiply(y2).subtract(y1.multiply(y1)));
BigDecimal t2 = (y3.subtract(y1)).multiply(x2.multiply(x2).subtract(x1.multiply(x1)));
BigDecimal t3 = (y1.subtract(y2)).multiply(y1.multiply(y1).subtract(y3.multiply(y3)));
BigDecimal t4 = (y1.subtract(y2)).multiply(x1.multiply(x1).subtract(x3.multiply(x3)));
BigDecimal t5 = BigDecimal.valueOf().multiply(y1.subtract(y2)).multiply(x3.subtract(x1));
BigDecimal t6 = BigDecimal.valueOf().multiply(y3.subtract(y1)).multiply(x1.subtract(x2));
BigDecimal x0 = (t1.add(t2).subtract(t3).subtract(t4)).divide(t5.subtract(t6)); BigDecimal v1 = y3.multiply(y3);
BigDecimal v2 = y1.multiply(y1);
BigDecimal v3 = BigDecimal.valueOf().multiply(x0).multiply(x3.subtract(x1));
BigDecimal v4 = x1.multiply(x1);
BigDecimal v5 = x3.multiply(x3);
BigDecimal v6 = BigDecimal.valueOf().multiply(y3.subtract(y1));
// System.out.println(v6);
BigDecimal y0 = (v1.subtract(v2).subtract(v3).subtract(v4).add(v5)).divide(v6); BigDecimal z1 = (y0.subtract(y2)).multiply(y0.subtract(y2));
BigDecimal z2 = (x0.subtract(x2)).multiply(x0.subtract(x2));
BigDecimal r = z1.add(z2); BigDecimal tmp1 = (x4.subtract(x0)).multiply(x4.subtract(x0));
BigDecimal tmp2 = (y4.subtract(y0)).multiply(y4.subtract(y0));
BigDecimal dis = tmp1.add(tmp2);
if(dis.compareTo(r) > )
{
System.out.println("Accepted");
}
else
{
System.out.println("Rejected");
} }
}
}

参考emmm

HDU 6206 Apple【计算几何+高精度Java】的更多相关文章

  1. HDU 6206 Apple (高精确度+JAVA BigDecimal)

    Problem Description Apple is Taotao's favourite fruit. In his backyard, there are three apple trees ...

  2. HDU 6206 Apple ( 高精度 && 计算几何 && 三点构圆求圆心半径 )

    题意 : 给出四个点,问你第四个点是否在前三个点构成的圆内,若在圆外输出"Accepted",否则输出"Rejected",题目保证前三个点不在一条直线上. 分 ...

  3. hdu 6206 : Apple 【计算几何 + 分数类】

    题目链接 比赛时C++上__float128都被卡精度,然后扔给队友用Java的BigDecimal过了 算法不多说,求三角形外心可以参考 维基百科 https://zh.wikipedia.org/ ...

  4. HDU 6206 Apple

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6206 判断给定一点是否在三角形外接圆内. 给定三角形三个顶点的坐标,如何求三角形的外心的坐标呢? 知乎 ...

  5. Apple严控Java太不人性化

    转自:http://www.cdtarena.com/javapx/201307/9115.html Apple为了在系统安全方面得到更好的声誉,对更容易造成系统漏洞的Java进行着严格的控制,并在自 ...

  6. HDU 4998 Rotate (计算几何)

    HDU 4998 Rotate (计算几何) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4998 Description Noting is more ...

  7. hdu 4643 GSM 计算几何 - 点线关系

    /* hdu 4643 GSM 计算几何 - 点线关系 N个城市,任意两个城市之间都有沿他们之间直线的铁路 M个基站 问从城市A到城市B需要切换几次基站 当从基站a切换到基站b时,切换的地点就是ab的 ...

  8. HDU 4925 Apple Tree(推理)

    HDU 4925 Apple Tree 题目链接 题意:给一个m*n矩阵种树,每一个位置能够选择种树或者施肥,假设种上去的位置就不能施肥,假设施肥则能让周围果树产量乘2.问最大收益 思路:推理得到肯定 ...

  9. hdu 5429 Geometric Progression 高精度浮点数(java版本)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5429 题意:给一段长度不超过100的每个数字(可以是浮点数)的长度不超过1000的序列,问这个序列是否 ...

随机推荐

  1. visio应用程序相关设置-选项-常规

    1.用户名称,可读写 ApplicationSettings.UserName m_Visio.Window.Application.Settings.UserName = "BEIJING ...

  2. 【BZOJ 3925】[Zjoi2015]地震后的幻想乡 期望概率dp+状态压缩+图论知识+组合数学

    神™题........ 这道题的提示......(用本苣蒻并不会的积分积出来的)并没有 没有什么卵用 ,所以你发现没有那个东西并不会 不影响你做题 ,然后你就可以推断出来你要求的是我们最晚挑到第几大的 ...

  3. myisam_sort_buffer_size vs sort_buffer_size

    Q: I am MySQL on server with 6GB RAM. I need to know what is the difference between myisam_sort_buff ...

  4. [uva 1350]数位dp+二分

    题目链接:https://vjudge.net/problem/38405 #include<bits/stdc++.h> using namespace std; ][]; ]; lon ...

  5. TCP之close_wait

    TCP之close_wait 浏览:3697次  出处信息 /* * @author: ahuaxuan * @date: 2010-4-30 */ 查看各状态连接数: netstat -n | aw ...

  6. Spring任务调度<task:scheduled-tasks>【含cron参数详解】 (转载)

    Spring内部有一个task是Spring自带的一个设定时间自动任务调度 task使用的时候很方便,但是他能做的东西不如quartz那么的多! 可以使用注解和配置两种方式,配置的方式如下 引入Spr ...

  7. Idea工具点滴积累

    方法参数提示: keymap里面搜索 quick document,默认是F2

  8. 【HDU5772】String Problem [网络流]

    String Problem Time Limit: 10 Sec  Memory Limit: 64 MB[Submit][Status][Discuss] Description Input Ou ...

  9. codevs1163访问艺术馆 树形dp

    算裸的树形dp吧 回来复习一波 #include<cstdio> #include<cstring> #include<algorithm> #include< ...

  10. 【mysql优化】语句优化

    1.int型子查询陷阱 如下两个表: mysql> desc user; +----------+-------------+------+-----+---------+-------+ | ...