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;
}