链接

题意甚是难懂!这是第二遍做这道题了,依旧无法理解题意,搜了下题意。。。

首先需要判断是不是为凸多边形。(从一个顶点走一遍即可,要注意顺逆时针,题目中没有指明)

其次看一下圆是不是能够放入多边形内。(首先判断一下圆心是否在圆内,然后枚举圆心到所有边的距离与半径r进行比较)

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 10000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
#define _sign(x) ((x) > eps?1:((x) < -eps ? 2 :0))
struct Point
{
double x,y;
Point(double x=,double y=):x(x),y(y) {}
}p[N],ch[N];
int top;
struct Circle
{
Point c;
double r;
//Circle(Point c,double r):c(c),r(r){}
Point point (double a)
{
return Point(c.x+cos(a)*r,c.y+sin(a)*r);
}
};
typedef Point pointt;
pointt operator + (Point a,Point b)
{
return Point(a.x+b.x,a.y+b.y);
}
pointt operator - (Point a,Point b)
{
return Point(a.x-b.x,a.y-b.y);
}
double cross(Point a,Point b)
{
return a.x*b.y-a.y*b.x;
}
double mul(Point p0,Point p1,Point p2)
{
return cross(p1-p0,p2-p0);
}
double dis(Point a)
{
return sqrt(a.x*a.x+a.y*a.y);
}
//精度判正负
int dcmp(double x)
{
if(fabs(x)<eps) return ;
else return x<?-:;
}
int Graham(int n)
{
int i,s[] = {,,};
double t;
for(i = ; i < n&&s[]|s[]; i ++)
{
t = mul(p[(i+)%n],p[(i+)%n],p[i]);
s[_sign(t)] = ;
}
return s[]|s[];
}
double distoline(Point p,Point a,Point b)
{
Point v1 = b-a,v2 = p-a;
return fabs(cross(v1,v2)/dis(v1));
}
int inside(Point po,int n)
{
int i,s[] = {,,};
double t;
for(i = ;i < n&&s[]|s[];i ++)
{
t = mul(p[(i+)%n],po,p[i]);
s[_sign(t)] = ;
}
return s[]|s[];
}
int main()
{
int n,i;
Circle cc;
while(scanf("%d",&n)!=EOF)
{
if(n<) break;
scanf("%lf%lf%lf",&cc.r,&cc.c.x,&cc.c.y);
for(i = ; i < n; i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
top = ;
int m = Graham(n);
if(!m)
{
puts("HOLE IS ILL-FORMED");
continue;
}
int flag = ;
if(!inside(cc.c,n))
{
puts("PEG WILL NOT FIT");
continue;
}
for(i = ; i < n; i ++)
{
if(dcmp(distoline(cc.c,p[i],p[i-])-cc.r)<)
{
flag = ;
break;
}
}
if(!flag)
puts("PEG WILL NOT FIT");
else
puts("PEG WILL FIT");
}
return ;
}

poj1584A Round Peg in a Ground Hole的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. sqlite加密

    一直使用sqlite来管理本地的数据,但是Xcode中的SDK中集成的sqlite是免费的,不提供加密模块,但是程序中用到的很多数据,有时候是不想让别人看到,一开始虑修改sqlite的源码,自己重新编 ...

  2. System.exit(0)和System.exit(1)区别:

    System.exit(0)是将你的整个虚拟机里的内容都停掉了,而finish()只是退出了activity,并没有退出应用,Application还是存在于内存中的,除非被系统回收.无论如何,内存都 ...

  3. c# UDP通信

    using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...

  4. 关于android中Bundle的使用

      1.Android using Bundle for sharing variables 注:android中使用Bundle来共享变量,下例中Activity1和Activity2通过bundl ...

  5. 20150812 Asp.net 父窗体获取子窗体的返回值,更新父窗体文本控件(应用)

    1. 父窗体 **************************************** using System;using System.Collections;using System.C ...

  6. ZOJ 3593 One Person Game

    One Person Game Time Limit: 2 Seconds      Memory Limit: 65536 KB There is an interesting and simple ...

  7. jquery easyui中文培训文档

    目  录 1.... Accordion(可折叠标签)... 2 1.1          实例... 2 1.2          参数... 3 2.... DateBox(日期框)... 4 2 ...

  8. FileInputstream的available()方法

    摘自:http://greemranqq.iteye.com/blog/2051487

  9. dubbo源码之三——dubbo重构

    dubbo版本:2.5.4 转自:http://javatar.iteye.com/blog/1041832

  10. 使用mysql数据库,插入数据出现问号(?)的问题,解决方法

    首先,我用的mysql数据库是5.7.12版本. 出现的问题: 1.插入数据显示错误,插入不成功,出现:Incorrect string value: '\xCD\xF5\xD5\xBC\xBE\xA ...