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POJ 3384 半平面交

2024-07-04 00:09:02   225人阅读

Feng Shui
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 4185 Accepted: 1279 Special Judge

Description

Feng shui is the ancient Chinese practice of placement and arrangement of space to achieve harmony with the environment. George has recently got interested in it, and now wants to apply it to his home and bring harmony to it.

There is a practice which says that bare floor is bad for living area since spiritual energy drains through it, so George purchased two similar round-shaped carpets (feng shui says that straight lines and sharp corners must be avoided). Unfortunately, he is unable to cover the floor entirely since the room has shape of a convex polygon. But he still wants to minimize the uncovered area by selecting the best placing for his carpets, and asks you to help.

You need to place two carpets in the room so that the total area covered by both carpets is maximal possible. The carpets may overlap, but they may not be cut or folded (including cutting or folding along the floor border) — feng shui tells to avoid straight lines.

Input

The first line of the input file contains two integer numbers n and r — the number of corners in George’s room (3 ≤ n ≤ 100) and the radius of the carpets (1 ≤r ≤ 1000, both carpets have the same radius). The following n lines contain two integersxi and yi each — coordinates of the i-th corner (?1000 ≤xi, yi ≤ 1000). Coordinates of all corners are different, and adjacent walls of the room are not collinear. The corners are listed in clockwise order.

Output

Write four numbers x1, y1, x2,y2 to the output file, where (x1, y1) and (x2,y2) denote the spots where carpet centers should be placed. Coordinates must be precise up to 4 digits after the decimal point.

If there are multiple optimal placements available, return any of them. The input data guarantees that at least one solution exists.

Sample Input

#15 2 -2 0 -5 3 0 8 7 3 5 0
#24 3 0 0 0 8 10 8 10 0

Sample Output

#1-2 3 3 2.5
#23 5 7 3

Hint

Source


已知一个多边形，给定两个半径为R的圆，求圆不碰到多边形边界，不允许切割，但是两个圆可以重合的情况下，覆盖多边形面积最大时两个圆心坐标。

把多边形每一条边同时向内推进R，求半平面交，求出来的就是圆心可以放置的区域，由于要使面积最大，所以找所有点中距离最远的两个点。

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/4 15:03:55
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 10000000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)<eps)return 0;
	return x>0?1:-1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
Point operator + (const Point &a,const Point &b){
	return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
	return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
	return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
	return Point(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 dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point  a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Length(Point a){
	return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
	return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
	return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
	return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
	return a/Length(a);
}
Point Normal(Point a){
	return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
	return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
	return Length(Cross(b-a,c-a));
}
struct Line{
	Point p,v;
	double ang;
	Line(){};
	Line(Point p,Point v):p(p),v(v){
		ang=atan2(v.y,v.x);
	}
	bool operator < (const Line &L) const {
		return ang<L.ang;
	}
};
bool OnLeft(const Line &L,const Point &p){
	return dcmp(Cross(L.v,p-L.p))>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
	Point u=p-q;
	double t=Cross(w,u)/Cross(v,w);
	return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
	return GetLineIntersection(a.p,a.v,b.p,b.v);
}
vector<Point> HPI(vector<Line> L){
	int n=L.size();
	sort(L.begin(),L.end());//将所有半平面按照极角排序。
	/*for(int i=0;i<n;i++){
		cout<<"han  "<<i<<" ";
		printf("%.2lf  %.2lf  %.2lf  %.2lf  %.2lf\n",L[i].p.x,L[i].p.y,L[i].v.x,L[i].v.y,L[i].ang);
	}*/
	int first,last;
	vector<Point> p(n);
	vector<Line> q(n);
	vector<Point> ans;
	q[first=last=0]=L[0];
	for(int i=1;i<n;i++){
		while(first<last&&!OnLeft(L[i],p[last-1]))last--;//删除顶部的半平面
		while(first<last&&!OnLeft(L[i],p[first]))first++;//删除底部的半平面
		q[++last]=L[i];//将当前的半平面假如双端队列顶部。
		if(fabs(Cross(q[last].v,q[last-1].v))<eps){//对于极角相同的，选择性保留一个。
			last--;
			if(OnLeft(q[last],L[i].p))q[last]=L[i];
		}
		if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);//计算队列顶部半平面交点。
	}
	while(first<last&&!OnLeft(q[first],p[last-1]))last--;//删除队列顶部的无用半平面。
	if(last-first<=1)return ans;//半平面退化
	p[last]=GetLineIntersection(q[last],q[first]);//计算队列顶部与首部的交点。
	for(int i=first;i<=last;i++)ans.push_back(p[i]);//将队列中的点复制。
	return ans;
}
double PolyArea(vector<Point> p){
	int n=p.size();
	double ans=0;
	for(int i=1;i<n-1;i++)
		ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
	return fabs(ans)/2;
}
Point pp[200],v[200],v2[200];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n;
	 double R;
	 while(~scanf("%d%lf",&n,&R)){
		for(int i=0;i<n;i++)scanf("%lf%lf",&pp[i].x,&pp[i].y);
		reverse(pp,pp+n);
	    for(int i=0;i<n;i++){
			v[i]=pp[(i+1)%n]-pp[i];
			v2[i]=Normal(v[i]);
		}	
	    vector<Line> L;
		for(int i=0;i<n;i++)
			L.push_back(Line(pp[i]+v2[i]*R,v[i]));
		vector<Point> ans=HPI(L);
		Point a,b;
		if(ans.size()==1)a=b=ans[0];
		double dd=-INF;
		for(int i=0;i<ans.size();i++)
			for(int j=i+1;j<ans.size();j++)
				if(Length(ans[i]-ans[j])>dd){
					dd=Length(ans[i]-ans[j]);
					a=ans[i];
					b=ans[j];
				}
		printf("%.4f %.4f %.4f %.4f\n",a.x,a.y,b.x,b.y);
	 }
     return 0;
}



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