poj A Round Peg in a Ground Hole
http://poj.org/problem?id=1584
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std; const int maxn=;
const double pi=acos(-1.0);
const double eps=10e-; int cmp(double x)
{
if(fabs(x)<eps) return ;
if(x>) return ;
return -;
} double sqr(double x)
{
return x*x;
} struct point
{
double x,y;
point(){}
point(double a,double b):x(a),y(b){}
bool operator <(const point &a)const
{
return (x<a.x)||(x==a.x&&y<a.y);
}
friend point operator -(const point &a,const point &b){
return point(a.x-b.x,a.y-b.y);
}
double norm(){
return sqrt(sqr(x)+sqr(y));
}
}p[maxn],ch[maxn]; struct line
{
point a,b;
line(){}
line(point x,point y):a(x),b(y){}
}; double det(point a,point b,point c)
{
return ((b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y));
} double cross(point a,point b,point c)
{
return ((b.x-a.x)*(c.y-b.y)-(c.x-b.x)*(b.y-a.y));
}
double det1(const point &a,const point &b)
{
return a.x*b.y-a.y*b.x;
} double dot(const point &a,const point &b)
{
return a.x*b.x+a.y*b.y;
} double dis(point a,point b)
{
return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
} double dis_point_segment(const point p,const point s,const point t)
{
if(cmp(dot(p-s,t-s))<) return (p-s).norm();
if(cmp(dot(p-t,s-t))<) return (p-t).norm();
return fabs(det1(s-p,t-p)/dis(s,t));
} bool pointonsegment(point p,point s,point t)
{
return cmp(det1(p-s,t-s))==&&cmp(dot(p-s,p-t))<=;
} int convex_hull(point *p,int n,point *ch)
{
sort(p,p+n);
int m=;
for(int i=; i<n; i++)
{
while(m>&&det(ch[m-],ch[m-],p[i])<=) m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-; i>=; i--)
{
while(m>k&&det(ch[m-],ch[m-],p[i])<=) m--;
ch[m++]=p[i];
}
if(n>) m--;
return m;
} bool convex_hull1(point *p,int n)
{
int flag=;
p[n]=p[];
for(int i=; i<=n; i++)
{
//printf("%lf%lf %lf%lf %lf%lf\n",p[i-2].x,p[i-2].y,p[i-1].x,p[i-1].y,p[i].x,p[i].y);
int t=cmp(cross(p[i-],p[i-],p[i]));
//printf("%d\n",t);
if(!flag) flag=t;
if(flag*t<) return false;
}
return true;
}
int point_in(point t,point *ch,int n)
{
int num=,d1,d2,k;
ch[n]=ch[];
for(int i=; i<n; i++)
{
if(pointonsegment(t,ch[i],ch[i+])) return ;
k=cmp(det1(ch[i+]-ch[i],t-ch[i]));
d1=cmp(ch[i].y-t.y);
d2=cmp(ch[i+].y-t.y);
if(k>&&d1<=&&d2>) num++;
if(k<&&d2<=&&d1>) num--;
}
return num!=;
}
int main()
{
int n;
double r,x,y;
//freopen("sb.txt","w",stdout);
while(scanf("%d",&n)!=EOF)
{
if(n<) break;
scanf("%lf%lf%lf",&r,&x,&y);
point t(x,y);
for(int i=; i<n; i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
} if(!convex_hull1(p,n))
{
printf("HOLE IS ILL-FORMED\n");
continue;
}
int cn=convex_hull(p,n,ch);
if(point_in(t,ch,cn))
{
double max1=dis_point_segment(t,ch[],ch[]);
for(int i=; i<cn+; i++)
{
max1=min(max1,dis_point_segment(t,ch[i-],ch[i]));
}
if(max1-r>=) printf("PEG WILL FIT\n");
else printf("PEG WILL NOT FIT\n");
}
else printf("PEG WILL NOT FIT\n");
}
return ;
}
poj A Round Peg in a Ground Hole的更多相关文章
- POJ 1518 A Round Peg in a Ground Hole【计算几何=_=你值得一虐】
链接: http://poj.org/problem?id=1584 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...
- POJ 1584 A Round Peg in a Ground Hole【计算几何=_=你值得一虐】
链接: http://poj.org/problem?id=1584 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...
- POJ 1584 A Round Peg in a Ground Hole 判断凸多边形 点到线段距离 点在多边形内
首先判断是不是凸多边形 然后判断圆是否在凸多边形内 不知道给出的点是顺时针还是逆时针,所以用判断是否在多边形内的模板,不用是否在凸多边形内的模板 POJ 1584 A Round Peg in a G ...
- A Round Peg in a Ground Hole(凸包应用POJ 1584)
A Round Peg in a Ground Hole Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5684 Accepte ...
- POJ 1584 A Round Peg in a Ground Hole(判断凸多边形,点到线段距离,点在多边形内)
A Round Peg in a Ground Hole Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4438 Acc ...
- POJ 1584 A Round Peg in a Ground Hole 判断凸多边形,判断点在凸多边形内
A Round Peg in a Ground Hole Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5456 Acc ...
- POJ 1584 A Round Peg in a Ground Hole[判断凸包 点在多边形内]
A Round Peg in a Ground Hole Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6682 Acc ...
- POJ 1584:A Round Peg in a Ground Hole
A Round Peg in a Ground Hole Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5741 Acc ...
- A Round Peg in a Ground Hole(判断是否是凸包,点是否在凸包内,圆与多边形的关系)
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4628 Accepted: 1434 Description The D ...
随机推荐
- bzoj4028: [HEOI2015]公约数数列
Description 设计一个数据结构. 给定一个正整数数列 a_0, a_1, ..., a_{n - 1},你需要支持以下两种操作: 1. MODIFY id x: 将 a_{id} 修改为 x ...
- 快速查询本机IP 分类: windows常用小技巧 2014-04-15 09:28 138人阅读 评论(0) 收藏
第一步: 点击windows建(屏幕左下方),在搜索程序和文件文本框内输入:cmd 第二步: 点击Enter建进入. 第三步: 输入:ipconfig即可. 版权声明:本文为博主原创文章,未 ...
- loadView,viewDidLoad等几种方法的调用总结
viewDidLoad 此方法只有当view从nib文件初始化的时候才被调用.viewDidLoad用于初始化,加载时用到的. loadView 此方法在控制器的view为nil的时候被调用.虽然经常 ...
- media screen 响应式布局(知识点)
一.什么是响应式布局? 响应式布局是Ethan Marcotte在2010年5月份提出的一个概念,简而言之,就是一个网站能够兼容多个终端--而不是为每个终端做一个特定的版本.这个概念是为解决移动互联网 ...
- SRM 207 Div II Level Two: RegularSeason,字符串操作(sstream),多关键字排序(操作符重载)
题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=2866&rd=5853 主要是要对字符串的操作要熟悉,熟 ...
- Linux命令 - 删除文件(夹)
1.删除文件夹 rm –rf /var/test 将会删除/var/test目录以及其下的所有文件.文件夹 2.删除文件 rm -f /var/test/test.txt 将会强制删除/var/tes ...
- Tips--怎么使用谷歌搜索
修改hosts即可: hosts在哪? windows下:C:\Windows\System32\drivers\etc 管理员身份打开,并将下载好的hosts文件内容,添加到原有的hosts文件末尾 ...
- Linq101-Grouping Operators
using System; using System.Collections.Generic; using System.Linq; namespace Linq101 { class Groupin ...
- JS 改变input 输入框样式
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <hea ...
- (转)php中__autoload()方法详解
转之--http://www.php100.com/html/php/lei/2013/0905/5267.html PHP在魔术函数__autoload()方法出现以前,如果你要在一个程序文件中实例 ...