POJ 3384

如果你理解上一题的把多边形各边向内推进R的含义，那么，这题就是水题了^-^

当各边都向内推进R时，所得交点即可作为半径为R的圆的圆心了，那么，枚举推进后多边形各点的距离，取最远两点，即为答案了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAXN=150;
const double eps=1e-8;
struct point {
	double x,y;
};
point pts[MAXN],p[MAXN],q[MAXN];
int n,ansCnt,curCnt; double r;

int DB(double d){
	if(d>eps) return 1;
	if(d<-eps) return -1;
	return 0;
}

void initial(){
	for(int i=1;i<=n;i++){
		p[i]=pts[i];
	}
	p[n+1]=p[1];
	p[0]=p[n];
	ansCnt=n;
}

double dist(point a,point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void getline(point x,point y,double &a,double &b,double &c){
    a = y.y - x.y;
    b = x.x - y.x;
    c = y.x * x.y - x.x * y.y;
}

point intersect(point x,point y,double a,double b,double c){
    double u = fabs(a * x.x + b * x.y + c);
    double v = fabs(a * y.x + b * y.y + c);
    point pt;
    pt.x=(x.x * v + y.x * u) / (u + v);
    pt.y=(x.y * v + y.y * u) / (u + v);
    return  pt;
}

void cut(double a,double b,double c){
	curCnt=0;
	for(int i=1;i<=ansCnt;i++){
		if(DB(a*p[i].x+b*p[i].y+c)>=0) q[++curCnt]=p[i];
		else{
			if(DB(a*p[i-1].x+b*p[i-1].y+c)>0) q[++curCnt]=intersect(p[i],p[i-1],a,b,c);
			if(DB(a*p[i+1].x+b*p[i+1].y+c)>0) q[++curCnt]=intersect(p[i],p[i+1],a,b,c);
		}
	}
	for(int i=1;i<=curCnt;i++)
	p[i]=q[i];
	p[curCnt+1]=p[1];
	p[0]=p[curCnt];
	ansCnt=curCnt;
}

void slove(double r){
	initial();
	for(int i=1;i<=n;i++){
		point tt,ta,tb;
		tt.x=pts[i+1].y-pts[i].y;
		tt.y=pts[i].x-pts[i+1].x;
		double k=r/sqrt(tt.x*tt.x+tt.y*tt.y);
		ta.x=pts[i].x+k*tt.x;
		ta.y=pts[i].y+k*tt.y;
		tb.x=pts[i+1].x+k*tt.x;
		tb.y=pts[i+1].y+k*tt.y;
		double a,b,c;
		getline(ta,tb,a,b,c);
		cut(a,b,c);
	}
}

int main(){
	while(scanf("%d%lf",&n,&r)!=EOF){
		for(int i=1;i<=n;i++)
		scanf("%lf%lf",&pts[i].x,&pts[i].y);
		pts[n+1]=pts[1];
		slove(r);
		point ans1,ans2; double dis=0;
		ans1=ans2=p[1];
		for(int i=1;i<=ansCnt;i++){
			for(int j=i+1;j<=ansCnt;j++){
				double tmp=dist(p[i],p[j]);
				if(tmp>dis){
					dis=tmp;
					ans1=p[i];
					ans2=p[j];
				}
			}
		}
		printf("%.4lf %.4lf %.4lf %.4lf\n",ans1.x,ans1.y,ans2.x,ans2.y);
	}
	return 0;
}