HDOJ 3694 Fermat Point in Quadrangle

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求四边形的费马点

费马点的定义：在平面上的到四边形四个顶点A，B，C，D的点

当ABCD为凸四边形的时候

可以证明M点是费马点，M是对角线的交点

当ABCD为凹四边形的时候

在ABCD四点中有某一个点为费马点

所以，这个点就是个计算几何的模板题啦~~~~~~

我们也不需要判断四边形的凹凸性

把六条线段算出来，判断是不是有交点就好，然后暴力枚举每一个可能性

#include<bits/stdc++.h>using namespace std;const double eps=1e-6;const double inf=1e20;const double pi=acos(-1.0);const int maxp=1010;int sgn(double x){    if (fabs(x)<eps) return 0;    if (x<0) return -1;    else return 1;}double sqr(double x){    return x*x;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y){        x=_x;        y=_y;    }    void input(){        scanf("%lf%lf",&x,&y);    }    void output(){        printf("%.2lf %.2lf\n",x,y);    }    double operator ^ (const Point &b) const{        return x*b.y-y*b.x;    }    double operator * (const Point &b) const{        return x*b.x+y*b.y;    }    double len2(){        return x*x+y*y;    }    double len(){        return sqrt(x*x+y*y);    }    double distance(Point p){        return hypot(x-p.x,y-p.y);    }    Point operator - (const Point &b) const{        return Point(x-b.x,y-b.y);    }    Point operator + (const Point &b) const{        return Point(x+b.x,y+b.y);    }    Point operator * (const double &k) const{        return Point(x*k,y*k);    }    Point operator / (const double &k) const{        return Point(x/k,y/k);    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e){        s=_s;        e=_e;    }    Line(Point p,double angle){        s=p;        if (sgn(angle-pi/2)==0) e=(s+Point(0,1));        else e=(s+Point(1,tan(angle)));    }    Line(double a,double b,double c){        if (sgn(a)==0){            s=Point(0,-c/b);            e=Point(1,-c/b);        }        else if (sgn(b)==0){            s=Point(-c/a,0);            e=Point(-c/a,1);        }        else{            s=Point(0,-c/b);            e=Point(1,(-c-a)/b);        }    }    void input(){        s.input();        e.input();    }    int segcrossseg(Line v){        int d1=sgn((e-s)^(v.s-s));        int d2=sgn((e-s)^(v.e-s));        int d3=sgn((v.e-v.s)^(s-v.s));        int d4=sgn((v.e-v.s)^(e-v.s));        if ((d1^d2)==-2 && ((d3^d4)==-2) ) return 2;        return ((d1==0&&sgn((v.s-s)*(v.s-e))<=0) ||                (d2==0&&sgn((v.e-s)*(v.e-e))<=0) ||                (d3==0&&sgn((s-v.s)*(s-v.e))<=0) ||                (d4==0&&sgn((e-v.s)*(e-v.e))<=0));    }    Point crosspoint(Line v){        double a1=(v.e-v.s)^(s-v.s);        double a2=(v.e-v.s)^(e-v.s);        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));    }};Point A,B,C,D,M;Line AB,CD,AC,BD,AD,BC;double ans;double calc(Point x){    return A.distance(x)+B.distance(x)+C.distance(x)+D.distance(x);}int main(){    //freopen("input.txt","r",stdin);    while(1){        A.input();B.input();C.input();D.input();        if (A.x==-1&&A.y==-1&&B.x==-1&&B.y==-1&&C.x==-1&&C.y==-1&&D.x==-1&&D.y==-1) break;        ans=inf;        ans=min(ans,calc(A));        ans=min(ans,calc(B));        ans=min(ans,calc(C));        ans=min(ans,calc(D));        AB=Line(A,B);        CD=Line(C,D);        if (AB.segcrossseg(CD)>0){            M=AB.crosspoint(CD);            ans=min(ans,calc(M));        }        AC=Line(A,C);        BD=Line(B,D);        if (AC.segcrossseg(BD)>0){            M=AC.crosspoint(BD);            ans=min(ans,calc(M));        }        AD=Line(A,D);        BC=Line(B,C);        if (AD.segcrossseg(BC)>0){            M=AD.crosspoint(BC);            ans=min(ans,calc(M));        }        printf("%.4lf\n",ans);    }    return 0;}

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