半平面交 求解最大内接圆

上图使用来描述直线向内推进r这一过程的ac=r,ae是推进前的直线上的两点，现在求c、f的坐标，

由上可得三角形ace  与三角形abd相似，所以很快便可求出ab 的距离，即x坐标的移动距离，延长de，

从f向其做垂线，连接fd，便又可以由三角形相似求得y坐标的移动距离

 

下面是两道半平面交向内推进r的例子

http://poj.org/problem?id=3525

http://poj.org/problem?id=3384

主要讲讲3384

先用3525的方法向内推进r求出一个内核，则可知这个核上的任何一点都可以成为圆心坐标，要使的两个圆的重叠面积尽可能小，

就要使他们的圆心坐标尽可能的远，素以可以选取内核上最远的两个点作为圆心

 

View Code

#include <cmath>
#include <cstring>
#include <cstdio>
#include<algorithm>
using namespace std;
const  double INF = 1000000000.0;
const  int maxn = 1010;
const double eps = 1e-8;
inline double sgn(double x) {return fabs(x)<eps?0:(x>0?1:-1);}
struct Point{
    double x,y;
    Point(double tx=0,double ty=0){x=tx;y=ty;}
    bool operator == (const Point& t) const {
        return sgn(x-t.x)==0 && sgn(y-t.y)==0;
    }
}p[maxn],st[maxn],pp[maxn],points[maxn];
struct Seg{Point s,e;};
double dist(Point a,Point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
double cross(Point a,Point b,Point c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}
bool outside(Seg seg,Point p){return cross(seg.s,seg.e,p)>eps;}//strictly outside the ploygon
bool inside(Seg seg,Point p){return cross(seg.s,seg.e,p)<-eps;}//strictly inside the ploygon
Point Intersect(Point p1, Point p2, Point p3, Point p4) {
    double a1, b1, c1, a2, b2, c2, d;Point p;
    a1 = p1.y - p2.y; b1 = p2.x - p1.x; c1 = p1.x*p2.y - p2.x*p1.y;
    a2 = p3.y - p4.y; b2 = p4.x - p3.x; c2 = p3.x*p4.y - p4.x*p3.y;
    d = a1*b2 - a2*b1;
    if ( fabs(d) < eps )    return false;
    p.x = (-c1*b2 + c2*b1) / d;
    p.y = (-a1*c2 + a2*c1) / d;
    return p;
}
int n,ban_tot;
void CUT(Seg seg){//p must be clockwise        
    int i,tot=0;
    for(i=1;i<=ban_tot;i++){
        if(!outside(seg,p[i]))  pp[++tot]=p[i];//p[i] :inside or on the boundary
        else {//p[i]: must be outside 
            if(inside(seg,p[i-1]))     pp[++tot]=Intersect(seg.s,seg.e,p[i],p[i-1]);//p[i-1] must be inside
            if(inside(seg,p[i+1]))     pp[++tot]=Intersect(seg.s,seg.e,p[i],p[i+1]);//p[i+1] must be inside
        }
    }ban_tot=tot;
    pp[tot+1]=pp[1];pp[0]=pp[tot];
    memcpy(p,pp,sizeof(pp));
}
void solve(double r){  //push length r inside
    memcpy(p,points,sizeof(points)); ban_tot=n;//don't forget initialize
    for(int i = 1; i <= n; ++i){  
        Point ta, tb, tt;  
        tt.x = points[i+1].y-points[i].y;  
        tt.y = points[i].x-points[i+1].x;  
        double k = r / sqrt(tt.x * tt.x + tt.y * tt.y);  
        tt.x = tt.x * k;  tt.y = tt.y * k;  
        ta.x = points[i].x + tt.x;   ta.y = points[i].y + tt.y;  
        tb.x = points[i+1].x + tt.x; tb.y = points[i+1].y + tt.y;  
        Seg seg;seg.s=ta,seg.e=tb;
        CUT(seg);  
    }  
}  
Seg ts[110];
int main(){
    int i;double r;
    scanf("%d%lf",&n,&r);
        for(i=1;i<=n;i++) scanf("%lf%lf",&points[i].x,&points[i].y);
        points[n+1]=points[1];ban_tot=n;points[0]=points[n];
        solve(r);//printf("tot=%d\n",ban_tot);
        double ans=0;
        int a=1,b=1;//printf("%.3lf %.4lf\n",p[1].x,p[1].y);
        for(i=1;i<=ban_tot;i++)
        {
            for(int j=i+1;j<=ban_tot;j++)
            {
                if(dist(p[i],p[j])>ans)
                {
                    a=i;
                    b=j;
                    ans=dist(p[i],p[j]);
                }
            }
        }
        printf("%.4lf %.4lf %.4lf %.4lf\n",p[a].x,p[a].y,p[b].x,p[b].y);
    return 0;
}