UVA 1298 - Triathlon

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题意：

铁人三项比赛分成连续的三段：游泳、自行车和赛跑。现在每个单项比赛的长度还没定，但已知各选手在每项比赛中的平均速度（假定该平均速度和赛程长度无关），所以你可以设计每项比赛的长度，让某个特定的选手获胜。你的任务是判断哪些选手有可能获得冠军（并列冠军不算）。

解法：

假设比赛送长度为1，其中游泳长度x，自行车长度y，赛跑长度1-x-y，则选手i打败选手j的条件是：

x/vi + y/ui + (1-x-y)/wi  <  x/vj + y/uj + (1-x-y)/wj

整理上式得到不等式：

(1/vj-1/wj-1/vi+1/wi)x + (1/uj-1/wj-1/ui+1/wi)y + 1/wj-1/wi > 0

对于每个选手i，都可以得到相对于选手j的如上不等式（不等式在二维坐标系是一个半平面）

于是问题转化为解n-1个格式为Ax+By+c<0的不等式，即求半平面交

如果半平面交非空，则Yes

注意这题精度要求较高

 

//大白p263#include <cmath>#include <cstdio>#include <cstring>#include <string>#include <queue>#include <functional>#include <set>#include <iostream>#include <vector>#include <algorithm>using namespace std;const double eps=1e-9;//精度const int INF=0x3f3f3f3f;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;typedef vector<Point> Polygon;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?false:true;}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;}double torad(double deg){return deg/180*PI;}//角度化为弧度Vector Rotate(Vector a,double rad){//向量逆时针旋转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* sol){//计算凸包sort(p,p+n);int m=0;for(int i=0;i<n;i++){while(m>1&&Cross(sol[m-1]-sol[m-2],p[i]-sol[m-2])<=0) m--;sol[m++]=p[i];}int k=m;for(int i=n-2;i>=0;i--){while(m>k&&Cross(sol[m-1]-sol[m-2],p[i]-sol[m-2])<=0) m--;sol[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){return dcmp(Cross(s.a-p,s.b-p))==0&&dcmp(Dot(s.a-p,s.b-p))<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(Point* p,int n){//多边形有向面积double area=0;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;}double rotating_calipers(Point *ch,int n)//旋转卡壳{int q=1;double ans=0;ch[n]=ch[0];for(int p=0;p<n;p++){while(Cross(ch[q+1]-ch[p+1],ch[p]-ch[p+1])>Cross(ch[q]-ch[p+1],ch[p]-ch[p+1]))q=(q+1)%n;ans=max(ans,max(Length(ch[p]-ch[q]),Length(ch[p+1]-ch[q+1])));}return ans;}Polygon CutPolygon(Polygon poly,Point a,Point b){//用a->b切割多边形 返回左侧Polygon newpoly;int n=poly.size();for(int i=0;i<n;i++){Point c=poly[i];Point d=poly[(i+1)%n];if(dcmp(Cross(b-a,c-a))>=0) newpoly.push_back(c);if(dcmp(Cross(b-a,c-d))!=0){Point ip=GetLineIntersection(a,b-a,c,d-c);if(isPointOnSegment(ip,Segment(c,d))) newpoly.push_back(ip);}}return newpoly;}int GetCircleCircleIntersection(Circle c1,Circle c2,Point& p1,Point& p2){//求两圆相交double d=Length(c1.c-c2.c);if(dcmp(d)==0){if(dcmp(c1.r-c2.r)==0) return -1;//两圆重合return 0;}if(dcmp(c1.r+c2.r-d)<0) return 0;if(dcmp(fabs(c1.r-c2.r)-d)>0) return 0;double a=Angle(c2.c-c1.c,Vector(1,0));double da=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));p1=c1.point(a-da);p2=c1.point(a+da);if(p1==p2) return 1;return 2;}bool isPointOnleft(Point p,Line L){return dcmp(Cross(L.v,p-L.p))>0;}//点在直线左边 线上不算int HalfplaneIntersection(Line *L,int n,Point* poly){//半平面交sort(L,L+n);int first,last;Point* p=new Point[n];Line* q=new Line[n];q[first=last=0]=L[0];for(int i=1;i<n;i++){while(first<last&&!isPointOnleft(p[last-1],L[i])) last--;while(first<last&&!isPointOnleft(p[first],L[i])) first++;q[++last]=L[i];if(dcmp(Cross(q[last].v,q[last-1].v))==0){last--;if(isPointOnleft(L[i].p,q[last])) q[last]=L[i];}if(first<last) p[last-1]=GetLineIntersection(q[last-1].p,q[last-1].v,q[last].p,q[last].v);}while(first<last&&!isPointOnleft(p[last-1],q[first])) last--;if(last-first<=1) return 0;p[last]=GetLineIntersection(q[last].p,q[last].v,q[first].p,q[first].v);int m=0;for(int i=first;i<=last;i++) poly[m++]=p[i];return m;}//两点式化为一般式A = b.y-a.y, B = a.x-b.x, C = -a.y*(B)-a.x*(A);//--------------------------------------//--------------------------------------//--------------------------------------//--------------------------------------//--------------------------------------Point poly[110];Line L[110];int v[110],u[110],w[110];int main(){int n;while(scanf("%d",&n)==1&&n){for(int i=0;i<n;i++)scanf("%d%d%d",v+i,u+i,w+i);for(int i=0;i<n;i++){int idx=0;double k=10000;bool ok=true;for(int j=0;j<n;j++)if(i!=j){if(v[i]<=v[j]&&u[i]<=u[j]&&w[i]<=w[j]){ok=false;break;}if(v[i]>=v[j]&&u[i]>=u[j]&&w[i]>=w[j]) continue;double a=(k/v[j]-k/w[j])-(k/v[i]-k/w[i]);double b=(k/u[j]-k/w[j])-(k/u[i]-k/w[i]);double c=k/w[j]-k/w[i];Point p;Vector v(b,-a);if(fabs(a)>fabs(b)) p=Point(-c/a,0);else p=Point(0,-c/b);L[idx++]=Line(p,v);}if(ok){L[idx++]=Line(Point(0,0),Vector(0,-1));L[idx++]=Line(Point(0,0),Vector(1,0));L[idx++]=Line(Point(0,1),Vector(-1,1));if(!HalfplaneIntersection(L,idx,poly)) ok=false;}if(ok) puts("Yes");else puts("No");}}return 0;}