题解 UVA1298 【Triathlon】

题目链接：Link

Solution

这题的主要思想就是数形结合。设比赛总长度为1，其中游泳长度为x，自行车长度为y，赛跑长度为 1-x-y，则选手i打败选手j（非并列）的条件是 \(\dfrac{x}{v_i}+\dfrac{y}{u_i}+\dfrac{1-x-y}{w_i}<\dfrac{x}{v_j}+\dfrac{y}{u_j}+\dfrac{1-x-y}{w_j}\)

可以把它整理成 Ax+By+C>0的形式，其中

A=$ (\dfrac{1}{v_j}-\dfrac{1}{w_j})-(\dfrac{1}{v_i}-\dfrac{1}{w_i}) $

B=$ (\dfrac{1}{u_j}-\dfrac{1}{w_j})-(\dfrac{1}{u_i}-\dfrac{1}{w_i}) $

C=$ \dfrac{1}{w_j}-\dfrac{1}{w_i} $

则得到的所有半平面的并就是合法的解，可以由此判断是否能夺冠。

贴代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
struct Point
{
	double x,y;
	Point(double x=0,double y=0):x(x),y(y){}
};
typedef Point Vector;

inline Vector operator+(const Vector &A,const Vector &B)
{ return Vector(A.x+B.x,A.y+B.y); }

inline Vector operator-(const Point &a,const Point &b)
{ return Vector(a.x-b.x,a.y-b.y); }

inline Vector operator*(const Vector &A,double p)
{ return Vector(A.x*p,A.y*p); }

inline Vector operator/(const Vector &A,double p)
{ return Vector(A.x/p,A.y/p); }

inline bool operator<(const Point &a,const Point &b)
{ return a.x<b.x||(a.x==b.x&&a.y<b.y); }

const double ops=1e-10;
inline int dcmp(double x)
{ return (x>0?x:-x)<ops?0:(x>0?1:-1); }

inline bool operator==(const Point &a,const Point &b)
{ return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0; }

inline double Dot(const Vector &A,const Vector &B)
{ return A.x*B.x+A.y*B.y; }

inline double Length(const Vector &A)
{ return sqrt(Dot(A,A)); }

inline double Cross(const Vector &A,const Vector &B)
{ return A.x*B.y-A.y*B.x; }

inline Vector Normal(const Vector &A)
{
	double L=Length(A);
	return Vector(-A.y/L,A.x/L);
}

struct Line
{
	Point P;
	Vector v;
	double ang;
	Line(Point P=Point(),Vector v=Vector()):P(P),v(v) { ang=atan2(v.y,v.x); }
	inline bool operator<(const Line &L) const
	{ return ang<L.ang; }
};

inline bool OnLeft(const Line &L,const Point &p)
{ return dcmp(Cross(L.v,p-L.P))>0; }

Point GetLineIntersection(const Line &a,const Line &b)
{
	Vector u=a.P-b.P;
	double t=Cross(b.v,u)/Cross(a.v,b.v);
	return a.P+a.v*t;
}

int HalfplaneIntersection(Line *L,int n)
{
	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&&!OnLeft(L[i],p[last-1])) last--;
		while(first<last&&!OnLeft(L[i],p[first])) first++;
		q[++last]=L[i];
		if(dcmp(Cross(q[last].v,q[last-1].v))==0)
		{
			last--;
			if(OnLeft(q[last],L[i].P)) q[last]=L[i];
		}
		if(first<last) p[last-1]=GetLineIntersection(q[last-1],q[last]);
	}
	while(first<last&&!OnLeft(q[first],p[last-1])) last--;
	if(last-first<=1) return 0;
	p[last]=GetLineIntersection(q[last],q[first]);
	
	return max(last-first+1,0);
}
const int maxn=110;
const double k=10000;
Point poly[maxn];
Line L[maxn];
int n;
int V[maxn],U[maxn],W[maxn];
int main()
{
#ifdef local
	freopen("pro.in","r",stdin);
#endif
	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 lc=0,ok=1;
			for(int j=0;j<n;j++)
				if(i!=j)
				{
					if(V[i]<=V[j]&&U[i]<=U[j]&&W[i]<=W[j]) { ok=0; 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[lc++]=Line(P,v);
				}
			if(ok)
			{
				L[lc++]=Line(Point(0,0),Vector(0,-1));
				L[lc++]=Line(Point(0,0),Vector(1,0));
				L[lc++]=Line(Point(0,1),Vector(-1,1));
				if(!HalfplaneIntersection(L,lc)) ok=0;
			}
			printf("%s\n",ok?"Yes":"No");
		}
	}
	return 0;
}