UVA 11437 Triangle Fun

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水题

可以证明PQR面积是ABC面积的1/7

详细证明见http://www.cnblogs.com/freezhan/archive/2012/11/11/2776472.html

或者求交点暴力计算也是可以的

 

//大白p263#include <cmath>#include <cstdio>#include <cstring>#include <set>#include <iostream>#include <vector>#include <algorithm>using namespace std;const double eps=1e-9;//精度const int INF=1<<29;const double PI=acos(-1.0);int dcmp(double x){//判断double等于0或。。。    if(fabs(x)<eps)return 0;else return x<0?-1:1;}struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y){}};typedef Point Vector;Vector operator+(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}//向量+向量=向量Vector operator-(Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}//点-点=向量Vector operator*(Vector a,double p){return Vector(a.x*p,a.y*p);}//向量*实数=向量Vector operator/(Vector a,double p){return Vector(a.x/p,a.y/p);}//向量/实数=向量bool operator<( const Point& A,const Point& B ){return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0);}bool operator==(const Point&a,const Point&b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}bool operator!=(const Point&a,const Point&b){return a==b;}struct Segment{    Point a,b;    Segment(){}    Segment(Point _a,Point _b){a=_a,b=_b;}    bool friend operator<(const Segment& p,const Segment& q){return p.a<q.a||(p.a==q.a&&p.b<q.b);}    bool friend operator==(const Segment& p,const Segment& q){return (p.a==q.a&&p.b==q.b)||(p.a==q.b&&p.b==q.a);}};struct Circle{    Point c;    double r;    Circle(){}    Circle(Point _c, double _r):c(_c),r(_r) {}    Point point(double a)const{return Point(c.x+cos(a)*r,c.y+sin(a)*r);}    bool friend operator<(const Circle& a,const Circle& b){return a.r<b.r;}};struct Line{    Point p;    Vector v;    double ang;    Line() {}    Line(const Point &_p, const Vector &_v):p(_p),v(_v){ang = atan2(v.y, v.x);}    bool operator<(const Line &L)const{return  ang < L.ang;}};double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}//|a|*|b|*cosθ 点积double Length(Vector a){return sqrt(Dot(a,a));}//|a| 向量长度double Angle(Vector a,Vector b){return acos(Dot(a,b)/Length(a)/Length(b));}//向量夹角θdouble Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}//叉积 向量围成的平行四边形的面积double Area2(Point a,Point b,Point c){return Cross(b-a,c-a);}//同上 参数为三个点double DegreeToRadius(double deg){return deg/180*PI;}double GetRerotateAngle(Vector a,Vector b){//向量a顺时针旋转theta度得到向量b的方向    double tempa=Angle(a,Vector(1,0));    if(a.y<0) tempa=2*PI-tempa;    double tempb=Angle(b,Vector(1,0));    if(b.y<0) tempb=2*PI-tempb;    if((tempa-tempb)>0) return tempa-tempb;    else return tempa-tempb+2*PI;}Vector Rotate(Vector a,double rad){//向量旋转rad弧度    return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));}Vector Normal(Vector a){//计算单位法线    double L=Length(a);    return Vector(-a.y/L,a.x/L);}Point GetLineProjection(Point p,Point a,Point b){//点在直线上的投影    Vector v=b-a;    return a+v*(Dot(v,p-a)/Dot(v,v));}Point GetLineIntersection(Point p,Vector v,Point q,Vector w){//求直线交点 有唯一交点时可用    Vector u=p-q;    double t=Cross(w,u)/Cross(v,w);    return p+v*t;}int ConvexHull(Point* p,int n,Point* ch){//计算凸包    sort(p,p+n);    int m=0;    for(int i=0;i<n;i++){        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;        ch[m++]=p[i];    }    int k=m;    for(int i=n-2;i>=0;i--){        while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;        ch[m++]=p[i];    }    if(n>0) m--;    return m;}double Heron(double a,double b,double c){//海伦公式    double p=(a+b+c)/2;    return sqrt(p*(p-a)*(p-b)*(p-c));}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){//线段规范相交判定    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);    double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}double CutConvex(const int n,Point* poly, const Point a,const Point b, vector<Point> result[3]){//有向直线a b 切割凸多边形    vector<Point> points;    Point p;    Point p1=a,p2=b;    int cur,pre;     result[0].clear();     result[1].clear();     result[2].clear();    if(n==0) return 0;    double tempcross;    tempcross=Cross(p2-p1,poly[0]-p1);    if(dcmp(tempcross)==0) pre=cur=2;    else if(tempcross>0) pre=cur=0;    else pre=cur=1;    for(int i=0;i<n;i++){        tempcross=Cross(p2-p1,poly[(i+1)%n]-p1);        if(dcmp(tempcross)==0) cur=2;        else if(tempcross>0) cur=0;        else cur=1;        if(cur==pre){            result[cur].push_back(poly[(i+1)%n]);         }        else{            p1=poly[i];             p2=poly[(i+1)%n];            p=GetLineIntersection(p1,p2-p1,a,b-a);            points.push_back(p);             result[pre].push_back(p);             result[cur].push_back(p);             result[cur].push_back(poly[(i+1)%n]);             pre=cur;        }    }    sort(points.begin(),points.end());    if(points.size()<2){        return 0;     }    else{        return Length(points.front()-points.back());    }}double DistanceToSegment(Point p,Segment s){//点到线段的距离    if(s.a==s.b) return Length(p-s.a);    Vector v1=s.b-s.a,v2=p-s.a,v3=p-s.b;    if(dcmp(Dot(v1,v2))<0) return Length(v2);    else if(dcmp(Dot(v1,v3))>0) return Length(v3);    else return fabs(Cross(v1,v2))/Length(v1);}bool isPointOnSegment(Point p,Segment s){//点在线段上    return dcmp(DistanceToSegment(p,s))==0;}int isPointInPolygon(Point p, Point* poly,int n){//点与多边形的位置关系    int wn=0;    for(int i=0;i<n;i++){        Point& p2=poly[(i+1)%n];        if(isPointOnSegment(p,Segment(poly[i],p2))) return -1;//点在边界上        int k=dcmp(Cross(p2-poly[i],p-poly[i]));        int d1=dcmp(poly[i].y-p.y);        int d2=dcmp(p2.y-p.y);        if(k>0&&d1<=0&&d2>0)wn++;        if(k<0&&d2<=0&&d1>0)wn--;    }    if(wn) return 1;//点在内部    else return 0;//点在外部}double PolygonArea(vector<Point> p){//多边形有向面积    double area=0;    int n=p.size();    for(int i=1;i<n-1;i++)        area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;}int GetLineCircleIntersection(Line L,Circle C,Point& p1,Point& p2){//圆与直线交点 返回交点个数    double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y-C.c.y;    double e = a*a + c*c, f = 2*(a*b+c*d), g = b*b + d*d -C.r*C.r;    double delta = f*f - 4*e*g;    if(dcmp(delta) < 0)  return 0;//相离    if(dcmp(delta) == 0) {//相切        p1=p1=C.point(-f/(2*e));        return 1;    }//相交    p1=(L.p+L.v*(-f-sqrt(delta))/(2*e));    p2=(L.p+L.v*(-f+sqrt(delta))/(2*e));    return 2;}//--------------------------------------//--------------------------------------//--------------------------------------//--------------------------------------//--------------------------------------int main(){int T,n;scanf("%d",&T);Point a,b,c;while(T--){scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);double s=Heron(Length(a-b),Length(a-c),Length(b-c));printf("%.0lf\n",s/7);}return 0;}