链接

一直理解的最小覆盖圆就是外接圆。。原来还要分钝角和锐角。。。

钝角的话就为最长边的中点,对于这题分别枚举一下外接圆以及中点的圆,判一下是不是在园外。

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
struct Point
{
double x,y;
Point (double x=,double y =):x(x),y(y){}
}p[];
struct Circle
{
Point center;
double r;
};
typedef Point pointt;
pointt operator - (Point a,Point b)
{
return Point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
if(fabs(x)<eps) return ;
return x<?-:;
}
double dis(Point a)
{
return a.x*a.x+a.y*a.y;
}
double cross(Point a,Point b)
{
return a.x*b.y-a.y*b.x;
}
double area()
{
return fabs(cross(p[]-p[],p[]-p[]))/;
}
struct Circle Circumcircle()
{
Circle tmp;
double a,b,c,c1,c2;
double xa,ya,xb,yb,xc,yc;
a = sqrt(dis(p[]-p[]));
b = sqrt(dis(p[]-p[]));
c = sqrt(dis(p[]-p[]));
//¸ù¾Ýs = a*b*c/R/4£¬Çó°ë¾¶
tmp.r = (a*b*c)/(area()*4.0);
xa = p[].x;
ya = p[].y;
xb = p[].x;
yb = p[].y;
xc = p[].x;
yc = p[].y;
c1 = (dis(p[])-dis(p[]))/;
c2 = (dis(p[])-dis(p[]))/;
tmp.center.x = (c1*(ya-yc)-c2*(ya-yb))/((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb));
tmp.center.y = (c1*(xa-xc)-c2*(xa-xb))/((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));
return tmp;
}
int main()
{
int t,i;
cin>>t;
int kk = ;
while(t--)
{
for(i = ;i <= ; i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
Circle cc = Circumcircle();
Point pp;
scanf("%lf%lf",&pp.x,&pp.y);
double r = cc.r;
r*=r;
printf("Case #%d: ",++kk);
if(dis(pp-cc.center)>r)
{
puts("Safe");
continue;
}
r = dis(p[]-p[])/;
cc.center.x = (p[].x+p[].x)/;
cc.center.y = (p[].y+p[].y)/;
if(dcmp(dis(p[]-cc.center)-r)<=&&dcmp(dis(pp-cc.center)-r)>)
{
puts("Safe");
continue;
}
r = dis(p[]-p[])/;
cc.center.x = (p[].x+p[].x)/;
cc.center.y = (p[].y+p[].y)/;
if(dcmp(dis(p[]-cc.center)-r)<=&&dcmp(dis(pp-cc.center)-r)>)
{
puts("Safe");
continue;
}
r = dis(p[]-p[])/;
cc.center.x = (p[].x+p[].x)/;
cc.center.y = (p[].y+p[].y)/;
if(dcmp(dis(p[]-cc.center)-r)<=&&dcmp(dis(pp-cc.center)-r)>)
{
puts("Safe");
continue;
}
puts("Danger");
}
return ;
}

hdu4720Naive and Silly Muggles的更多相关文章

  1. 计算几何 HDOJ 4720 Naive and Silly Muggles

    题目传送门 /* 题意:给三个点求它们的外接圆,判断一个点是否在园内 计算几何:我用重心当圆心竟然AC了,数据真水:) 正解以后补充,http://www.cnblogs.com/kuangbin/a ...

  2. Naive and Silly Muggles

    Problem Description Three wizards are doing a experiment. To avoid from bothering, a special magic i ...

  3. Naive and Silly Muggles (计算几何)

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/ ...

  4. HDU 4720 Naive and Silly Muggles (外切圆心)

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...

  5. HDU 4720Naive and Silly Muggles热身赛2 1005题(分锐角钝角三角形讨论)

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/ ...

  6. Naive and Silly Muggles hdu4720

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/ ...

  7. HDU 4720 Naive and Silly Muggles (简单计算几何)

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/ ...

  8. HDUOJ-------Naive and Silly Muggles

    Naive and Silly Muggles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/ ...

  9. ACM学习历程—HDU4720 Naive and Silly Muggles(计算几何)

    Description Three wizards are doing a experiment. To avoid from bothering, a special magic is set ar ...

随机推荐

  1. libsqlite3.dylib找不到

    Xcode7中 Link Binary With Libraries 没有 .dylib库,只能找到对应的 .tbd,但不能代替使用,通过查找资料,尝试后得到以下两种解决方法. 方法1. (heqin ...

  2. 荒木毬菜 小情歌日文版 - 独身OL之歌

    咎(とが)めるつもりもないけどtogameru tumorimo naikedo并不想责备在身旁 暇(ひま)してる时间(じかん)をhimashiteru jikan wo无所事事的时间 パジャマの鸟( ...

  3. PostgreSQL中使用外部表

    1. 安装file_fdw 需要先安装file_fdw,一般是进到PostgreSQL的源码包中的contrib/file_fdw目录下,执行: make make install 然后进入数据库中, ...

  4. java 判断某一天是当年的哪一天

    题目:输入年份,月份,日,判断这一天是这一年的第几天?(闰年的2月份为29天,平年为28天) public class Runnian { /** * 能被4整除且不能被100整除或者能被400整除的 ...

  5. [算法][C]计算向量的角度

    C 语言里 double atan2(double y,double x) 返回的是原点至点(x,y)的方位角,即与 x 轴的夹角.也可以理解为复数 x+yi 的辐角.返回值的单位为弧度,取值范围为 ...

  6. 获取SqlServer2005表结构(字段,主键,外键,递增,描述)

    1.获取表的基本字段属性 --获取SqlServer中表结构 SELECT syscolumns.name,systypes.name,syscolumns.isnullable, syscolumn ...

  7. 浅谈算法和数据结构: 七 二叉查找树 八 平衡查找树之2-3树 九 平衡查找树之红黑树 十 平衡查找树之B树

    http://www.cnblogs.com/yangecnu/p/Introduce-Binary-Search-Tree.html 前文介绍了符号表的两种实现,无序链表和有序数组,无序链表在插入的 ...

  8. CentOS 修改线程数限制等(limits.conf)

    修改/etc/security/limits.conf,例如启动程序的用户为webadmin,则添加以下配置: webadmin - nofile 65536 webadmin - nproc 655 ...

  9. 华为S5700S配置总结

    需要使用通讯控制线缆连接电脑和交换机, 一头接交换机的Console口,一头接电脑(应该需要串口,没有的话得用USB转串口). 在PC机上运行终端仿真程序,设置终端通信参数为: 波特率为9600bit ...

  10. 为什么需要main函数,及其参数的用法

    首先,需要明确main函数是什么? 答:main函数是C语言约定的入口函数 C99标准里面是这样描述的: Program startup The function called at program ...