URAL 1062 - Triathlon（半平面交）
这个题乍眼一看好像很简单，然后我就认为u、v、w只要有全部比另外一个人小的就不能win，否则就能win，但是这个思路只对了一半
不能win的结论是正确的，但是win的结论不止排除这一个条件
将这个人与其他人的条件列式
如果都win的话，则满足 x/v+y/u+(k-x-y)/w（i的）<x/v+y/u+(k-x-y)/w（j的）由此构成n条直线（半平面），然后求交，如果有交集，则输出Yes，否则No
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y) {}
};
typedef Point Vector;
Vector operator - (Point A, Point B)
{
return Vector(A.x-B.x, A.y-B.y);
}
struct Line
{
Point p;
Vector v;
double ang;
Line() {}
Line(Point p,Vector v):p(p),v(v)
{
ang=atan2(v.y,v.x);
}
bool operator < (const Line& L) const
{
return ang<L.ang;
}
};
double Cross(Vector A, Vector B)
{
return A.x*B.y - A.y*B.x;
}
bool Onleft(Line L, Point p)
{
return Cross(L.v, p-L.p)>0;
}
Point GetIntersection(Line a, Line b)
{
Vector u = a.p-b.p;
double t = Cross(b.v, u)/Cross(a.v, b.v);
return Point(a.p.x+a.v.x*t, a.p.y+a.v.y*t);
}
int BPMJ(Line *L, int n, Point *poly)
{
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(fabs(Cross(q[last].v,q[last-1].v))<eps)
{
last--;
if(Onleft(q[last], L[i].p))q[last] = L[i];
}
if(first<last) p[last-1] = GetIntersection(q[last-1], q[last]);
}
while(first<last && !Onleft(q[first], p[last-1])) last--;
if(last-first<=1)return 0;
p[last] = GetIntersection(q[last], q[first]);
int m = 0;
for(int i=first; i<=last; i++)
poly[m++] = p[i];
return m;
}
double k=10000.0;
int u[110],v[110],w[110];
Point poly[110];
Line L[110];
int main()
{
int n;
while(scanf("%d",&n)==1)
{
for(int i=0; i<n; i++)
{
scanf("%d%d%d",&u[i],&v[i],&w[i]);
}
for(int i=0; i<n; i++)
{
int f=0;
for(int j=0; j<n; j++)
{
if(j!=i && u[i]<=u[j] && v[i]<=v[j] && w[i]<=w[j])
{
f=1;
break;
}
}
if(f==1)
printf("No\n");
else
{
Point p;
int ct=0;
for(int j=0; j<n; j++)
{
if(j!=i && u[i]>=u[j] && v[i]>=v[j] && w[i]>=w[j])
continue;
else if(j!=i)
{
double A=k/v[j]-k/v[i]-k/w[j]+k/w[i];
double B=k/u[j]-k/u[i]-k/w[j]+k/w[i];
double C=k/w[j]-k/w[i];
if(fabs(A)>fabs(B))
p=Point(-C/A,0);
else
p=Point(0,-C/B);
Vector v(B,-A);
L[ct++] = Line(p,v);
}
}
L[ct++] = Line(Point(0,0), Vector(0,-1));
L[ct++] = Line(Point(0,0), Vector(1,0));
L[ct++] = Line(Point(0,1), Vector(-1,1));
if(BPMJ(L,ct,poly))printf("Yes\n");
else printf("No\n");
}
}
}
return 0;
}