POJ 1584 A Round Peg in a Ground Hole --判定点在形内形外形上
题意: 给一个圆和一个多边形,多边形点可能按顺时针给出,也可能按逆时针给出,先判断多边形是否为凸包,再判断圆是否在凸包内。
解法: 先判是否为凸包,沿着i=0~n,先得出初始方向dir,dir=1为逆时针,dir=-1为顺时针,然后如果后面有两个相邻的边叉积后得出旋转方向为nowdir,如果dir*nowdir < 0,说明方向逆转了,即出现了凹点,说明不是凸多边形。
然后判圆是否在多边形内: 先判圆心是否在多边形内,用环顾法,然后如果在之内,则依次判断圆心与每条凸包边的距离与半径的距离,如果所有的dis都大于等于R,说明圆在凸包内。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define pi acos(-1.0)
#define eps 1e-8
using namespace std; struct Point{
double x,y;
Point(double x=, double y=):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
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); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -;
if(x > eps) return ;
return ;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == && dcmp(a.y-b.y) == ; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < ) return Length(v2);
if(dcmp(Dot(v1, v3)) > ) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
//点是否在多边形内部
int CheckPointInPolygon(Point A,Point* p,int n){
double TotalAngle = 0.0;
for(int i=;i<n;i++) {
if(dcmp(Cross(p[i]-A,p[(i+)%n]-A)) >= ) TotalAngle += Angle(p[i]-A,p[(i+)%n]-A);
else TotalAngle -= Angle(p[i]-A,p[(i+)%n]-A);
}
if(dcmp(TotalAngle) == ) return ; //外部
else if(dcmp(fabs(TotalAngle)-*pi) == ) return ; //完全内部
else if(dcmp(fabs(TotalAngle)-pi) == ) return ; //边界上
else return ; //多边形顶点
}
//判断未知时针方向的多边形是否是凸包
bool CheckConvexHull(Point* p,int n){
int dir = ; //旋转方向
for(int i=;i<n;i++) {
int nowdir = dcmp(Cross(p[(i+)%n]-p[i],p[(i+)%n]-p[i]));
if(!dir) dir = nowdir;
if(dir*nowdir < ) return false; //非凸包
}
return true;
} Point p[]; int main()
{
int n,i,j;
Circle Peg;
while(scanf("%d",&n)!=EOF && n >= )
{
scanf("%lf",&Peg.r); Peg.c.input();
for(i=;i<n;i++) p[i].input();
if(!CheckConvexHull(p,n)) { puts("HOLE IS ILL-FORMED"); continue; }
if(CheckPointInPolygon(Peg.c,p,n))
{
for(i=;i<n;i++)
{
double dis = DistanceToSeg(Peg.c,p[i],p[(i+)%n]);
if(dcmp(dis-Peg.r) < ) break;
}
if(i == n) { puts("PEG WILL FIT"); continue; }
}
puts("PEG WILL NOT FIT");
}
return ;
}
参考文章: http://blog.csdn.net/lyy289065406/article/details/6648606
射线法:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=, double y=):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
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); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -;
if(x > eps) return ;
return ;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == && dcmp(a.y-b.y) == ; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); } bool OnSegment(Point P, Point A, Point B) {
return dcmp(Cross(A-P,B-P)) == && dcmp(Dot(A-P,B-P)) <= ;
}
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < ) return Length(v2);
if(dcmp(Dot(v1, v3)) > ) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
//判断未知时针方向的多边形是否是凸包
bool CheckConvexHull(Point* p,int n){
int dir = ; //旋转方向
for(int i=;i<n;i++) {
int nowdir = dcmp(Cross(p[(i+)%n]-p[i],p[(i+)%n]-p[i]));
if(!dir) dir = nowdir;
if(dir*nowdir < ) return false; //非凸包
}
return true;
}
int Ray_PointInPolygon(Point A,Point* p,int n) {
int wn = ;
for(int i=;i<n;i++) {
//if(OnSegment(A,p[i],p[(i+1)%n])) return -1; //边界
int k = dcmp(Cross(p[(i+)%n]-p[i], A-p[i]));
int d1 = dcmp(p[i].y-A.y);
int d2 = dcmp(p[(i+)%n].y-A.y);
if(k > && d1 <= && d2 > ) wn++;
if(k < && d2 <= && d1 > ) wn--;
}
if(wn) return ; //内部
return ; //外部
} Point p[]; int main()
{
int n,i,j;
Circle Peg;
while(scanf("%d",&n)!=EOF && n >= )
{
scanf("%lf",&Peg.r); Peg.c.input();
for(i=;i<n;i++) p[i].input();
if(!CheckConvexHull(p,n)) { puts("HOLE IS ILL-FORMED"); continue; }
if(Ray_PointInPolygon(Peg.c,p,n))
{
for(i=;i<n;i++)
{
double dis = DistanceToSeg(Peg.c,p[i],p[(i+)%n]);
if(dcmp(dis-Peg.r) < ) break;
}
if(i == n) { puts("PEG WILL FIT"); continue; }
}
puts("PEG WILL NOT FIT");
}
return ;
}
POJ 1584 A Round Peg in a Ground Hole --判定点在形内形外形上的更多相关文章
- 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 ...
- 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)
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(判断凸多边形,点到线段距离,点在多边形内)
http://poj.org/problem?id=1584 题意 按照顺时针或逆时针方向输入一个n边形的顶点坐标集,先判断这个n边形是否为凸包. 再给定一个圆形(圆心坐标和半径),判断这个圆是否完全 ...
- 简单几何(点的位置) POJ 1584 A Round Peg in a Ground Hole
题目传送门 题意:判断给定的多边形是否为凸的,peg(pig?)是否在多边形内,且以其为圆心的圆不超出多边形(擦着边也不行). 分析:判断凸多边形就用凸包,看看点集的个数是否为n.在多边形内用叉积方向 ...
- POJ 1584 A Round Peg in a Ground Hole
先判断是不是N多边形,求一下凸包,如果所有点都用上了,那么就是凸多边形 判断圆是否在多边形内, 先排除圆心在多边形外的情况 剩下的情况可以利用圆心到每条边的最短距离与半径的大小来判断 #include ...
- 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 ...
随机推荐
- jquery实现输入框实时输入触发事件代码
$('.aa').bind('input propertychange', function() { searchProductClassbyName(); }); function searchPr ...
- PyInstaller编译python3时使用的详细参数介绍
继续翻译中.... The syntax of the pyinstaller command is: pyinstaller [options] script [script ...] | spec ...
- Ubuntu开机黑屏,无法进入系统
今天早上起来开机发现Ubuntu进不去了,启动项选择之后长时间的black of screen,击键盘.点鼠标毫无反应,后来实在等不下去了就按了一下电源键,以平时的性格就是强制关机的,这次轻轻碰一下就 ...
- VMware: XXX is still busy. Please wait until the operation is complete before closing
在使用vmware的过程中发现创建快照.恢复快照.管理快照等功能突然都变成灰色的,用不了.更觉得夸张的是仅仅剩下关闭虚机按钮是红色的.心想估计是虚机快照没处理完之类的问题导致的,于是想想关闭虚机重 ...
- RxJava 和 RxAndroid 四(RxBinding的使用)
对Rxjava不熟悉的同学可以先看我之前写的几篇文章 RxJava 和 RxAndroid 一 (基础) RxJava 和 RxAndroid 二(操作符的使用) RxJava 和 RxAndroid ...
- Android每次运行项目时重新启动一个新的模拟器的解决办法
具体解决办法 1.打开任务管理器,结束adb进程 2.此时android console下面会出现错误信息 3.切换到dos下面运行: adb start-server 4.重新运行android项目 ...
- iOS 获取emoji表情和拦截emoji表情
1 2 //将数字转为 #define EMOJI_CODE_TO_SYMBOL(x) ((((0x808080F0 | (x & 0x3F000) >> 4) | (x &a ...
- PagerTabStrip在ViewPager的页面中添加标题显示
package com.qf.day18_viewpager_demo_05; import java.util.ArrayList; import java.util.List; import an ...
- 【读书笔记】iOS-iCloud编程
一,苹果云服务-iCloud. 苹果公司斥资10亿美元在北卡罗来纳州简历数所中心-iDataCenter,该数据中心面积为50万平方英尺,也是美国最大规模的数据中心之一. 二,配置iCloud. 1, ...
- iOS 学习 - 4.存储聊天记录
主要是用sqlite3来存储聊天记录 先导入sqlite3.dylib, 点 Add Other,同时按住shift+command+G, 在弹出的Go to the folder中输入/usr/li ...