<題目鏈接>

題目大意：

給出一個凸多邊形的房間，根據風水要求，把兩個圓形地毯鋪在房間裏，不能折疊，不能切割，可以重疊。問最多能覆蓋多大空間，輸出兩個地毯的圓心坐標。多組解輸出其中一個，題目保證至少可以放入一個圓。

解題分析：

因為放置的圓不能超出多邊形的邊界，所以先將該凸多邊形的各個邊長向內平移 r 的距離，然後對這些平移後的直線用半平面交去切割原多邊形，切割後得到的區域就是兩圓圓心所在的區域，然後遍歷這個切割後的多邊形的各個頂點，距離最遠的兩個頂點就是這兩圓的圓心。

#include<algorithm>
#include<iostream>
#include
 
<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 10005

const double pi=acos(-1.0);
const double inf=1e9;
const double eps=1e-12;
int dcmp(double x)
{
    if (x<=eps&&x>=-eps) return 0;
    return (x>0)?1:-1;
}
struct Vector
{
    double x,y;
    Vector(
 
double X=0,double Y=0)
    {
        x=X,y=Y;
    }
};
typedef Vector Point;
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);}

int n,cnt,ncnt,ansi,ansj;
 
double x,y,r,ans;
Point p[N],poly[N],npoly[N];

double Dot(Vector a,Vector b)
{
    return a.x*b.x+a.y*b.y;
}
double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}
double Len(Vector a)
{
    return sqrt(Dot(a,a));
}
Vector rotate(Vector a,double rad)
{
    return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
bool insLS(Point A,Point B,Point C,Point D)
{
    Vector v=B-A,w=C-A,u=D-A;
    return dcmp(Cross(v,w))!=dcmp(Cross(v,u));
}
Point GLI(Point P,Vector v,Point Q,Vector w)
{
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
void init()
{
    cnt=0;
    poly[++cnt]=Point(inf,inf);
    poly[++cnt]=Point(inf,-inf);
    poly[++cnt]=Point(-inf,-inf);
    poly[++cnt]=Point(-inf,inf);
}
void halfp(Point A,Point B)
{
    ncnt=0;
    Point C,D;
    for (int i=1;i<=cnt;++i)
    {
        C=poly[i%cnt+1];
        D=poly[(i+1)%cnt+1];
        if (dcmp(Cross(B-A,C-A))<=0)
            npoly[++ncnt]=C;
        if (insLS(A,B,C,D))
            npoly[++ncnt]=GLI(A,B-A,C,D-C);
    }
    cnt=ncnt;
    for (int i=1;i<=cnt;++i)
        poly[i]=npoly[i];
}
int main()
{
    scanf("%d%lf",&n,&r);
    for (int i=1;i<=n;++i)
    {
        scanf("%lf%lf",&x,&y);
        p[i]=Point(x,y);
    }
    init();
    for (int i=1;i<=n;++i)
    {
        Point A=p[i%n+1];
        Point B=p[(i+1)%n+1];
        Vector v=B-A;
        Vector w=rotate(v,-pi/2.0);
        w=w*(r/Len(w));
        Point C=A+w;
        Point D=C+v;
        halfp(C,D);
    }
    ansi=ansj=1;
    for (int i=1;i<=cnt;++i)
        for (int j=i+1;j<=cnt;++j)
        {
            double dis=Len(poly[i]-poly[j]);
            if (dcmp(dis-ans)>0)
            {
                ans=dis;
                ansi=i,ansj=j;
            }
        }
    printf("%.4lf %.4lf %.4lf %.4lf\n",poly[ansi].x,poly[ansi].y,poly[ansj].x,poly[ansj].y);
}

2018-08-03

POJ 3384 放地毯【半平面交】