hdu 3694 费马点的应用

题目任意给出四个点，求出到这四个点的距离之和最小的点到这四个点的距离

若四边形为凸的，费马点为对角线交点，否则为凹的那点

证明很简单，把要证明的那点与其他顶点连起来，再任取一点，证明这点到四个顶点的距离比原来那点长即可

#include<stdio.h>
#include<math.h>
#include<string.h>
const double esp=1e-8;
struct point{
    double x,y;
}a[10];
struct line 
{
    double a,b,c;
};
double min(double a,double b)
{
    return a<b?a:b;
}
double dis(point a,point b)
{
    return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));
}
int dblcmp(double d) 
{
    if(fabs(d)<esp)
        return 0;
    return (d>0)?1:-1;
}
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);
}
void lineintersect(line l1,line l2, double &x, double &y)   
{
    double d=l1.a*l2.b-l2.a*l1.b;
     x=(l2.c*l1.b-l1.c*l2.b)/d;
     y=(l2.a*l1.c-l1.a*l2.c)/d;
}
void lineform(double x1,double y1,double x2,double y2, line &temp)//ax+by+c=0;
{
     temp.a=y2-y1;
     temp.b=x1-x2;
     temp.c=x2*y1-x1*y2;
}
double dit[5];
int main()
{
    int i,j;
    double ans;
    point fermat;
    while(1)
    {
        bool flag=false;
        for(i=1;i<=4;i++)
        {
            scanf("%lf%lf",&a[i].x,&a[i].y);
            if(a[i].x!=-1||a[i].y!=-1)
                flag=true;
        }
        if(!flag)
            break;
        memset(dit,0,sizeof(dit));
        for(i=1;i<=4;i++)
        {
            for(j=1;j<=4;j++)
            {
                dit[i]+=dis(a[i],a[j]);
            }
        }
        line temp1,temp2;
        double fx,fy;
        if(cross(a[1],a[2],a[3])*cross(a[1],a[2],a[4])<0)
        {
               lineform(a[1].x,a[1].y,a[2].x,a[2].y,temp1);
               lineform(a[3].x,a[3].y,a[4].x,a[4].y,temp2);
               lineintersect(temp1,temp2,fx,fy);
        }
        else if(cross(a[1],a[3],a[4])*cross(a[1],a[3],a[2])<0)
        {
               lineform(a[1].x,a[1].y,a[3].x,a[3].y,temp1);
               lineform(a[2].x,a[2].y,a[4].x,a[4].y,temp2);
               lineintersect(temp1,temp2,fx,fy);
        }
        else if(cross(a[1],a[4],a[3])*cross(a[1],a[4],a[2])<0)
        {
               lineform(a[1].x,a[1].y,a[4].x,a[4].y,temp1);
               lineform(a[2].x,a[2].y,a[3].x,a[3].y,temp2);
               lineintersect(temp1,temp2,fx,fy);
        }
        ans=0;
        fermat.x=fx;fermat.y=fy;
        for(i=1;i<=4;i++)
        {
           ans+=dis(fermat,a[i]);
        }
        double tans=min(min(dit[1],dit[2]),min(dit[3],dit[4]));
        double ret=min(tans,ans);
        printf("%.4lf\n",ret);
    }
    return 0;
}