HDU 4353 Finding Mine

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4353

题意：给出n个点，m个点，在n个点中选择一个多边形，其中包含的m个点中的点的个数为X，多边形面积为Y使得Y/X最小？

思路：多边形为三角形。

int DB(double x)

{

    if(x>EPS) return 1;

    if(x<-EPS) return -1;

    return 0;

}



class point

{

public:

    int x,y;



    point(){}

    point(double _x,double _y)

    {

        x=_x;

        y=_y;

    }



    point operator+(point a)

    {

        return point(x+a.x,y+a.y);

    }



    point operator-(point a)

    {

        return point(x-a.x,y-a.y);

    }



    int operator*(point a)

    {

        return x*a.y-y*a.x;

    }



    point operator*(double t)

    {

        return point(x*t,y*t);

    }



    double operator^(point a)

    {

        return x*a.x+y*a.y;

    }



    point operator/(double t)

    {

        return point(x/t,y/t);

    }



    double getLen()

    {

        return sqrt(x*x+y*y);

    }



    void get()

    {

        RD(x,y);

    }



    int operator==(point a)

    {

        return DB(x-a.x)==0&&DB(y-a.y)==0;

    }



    int operator<(point a)

    {

        if(DB(x-a.x)) return x<a.x;

        return y<a.y;

    }



    point adjust(double L)

    {

        double temp=getLen();

        return point(x*L/temp,y*L/temp);

    }



    void print()

    {

        printf("%.6lf %.6lf\n",x,y);

    }

};



int C,num;



point a[205],b[505];

int n,m,p[205][205];



int cross(point a,point b,point p)

{

    return (b-a)*(p-a);

}



int OK(point a,point b,point p)

{

    return p.x>=a.x&&p.x<b.x&&cross(a,b,p)>0;

}





double area(point a,point b,point c)

{

    return (a*b+b*c+c*a)/2.0;

}



int cmp(point a,point b)

{

    return a.x<b.x;

}



int main()

{

    RD(C);

    while(C--)

    {

        RD(n,m);

        int i,j,k,cnt;

        FOR1(i,n) a[i].get();

        FOR1(i,m) b[i].get();

        clr(p,0);

        sort(a+1,a+n+1,cmp);

        FOR1(i,n) for(j=i+1;j<=n;j++)

        {

            FOR1(k,m) if(OK(a[i],a[j],b[k]))

            {

                p[i][j]++;

            }

            p[j][i]=p[i][j];

        }

        double ans=1e20,temp;

        for(i=1;i<=n;i++) for(j=i+1;j<=n;j++) for(k=j+1;k<=n;k++)

        {

            cnt=abs(p[i][k]-p[j][k]-p[j][i]);

            if(cnt==0) continue;

            temp=fabs(area(a[i],a[j],a[k]))/cnt;

            if(temp<ans) ans=temp;

        }

        printf("Case #%d: ",++num);

        if(ans>1e10) puts("-1");

        else printf("%.6lf\n",ans);

    }

    return 0;

}