LA2218半平面交

没有想到这个题可以转化成解方程组的形式，就像线性规划一样，觉得好神奇。《训练指南》上的解析挺详细的，就不写了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=110;
const double eps=1e-8;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
};
typedef Point Vector;
struct DLine//有向直线directed line
{
    Point P;
    Vector v;
    double ang;
    DLine() {}
    DLine(Point P,Vector v):P(P),v(v)
    {
        ang=atan2(v.y,v.x);
    }
    bool operator < (const DLine& L) const
    {
        return ang<L.ang;
    }
};

Vector operator + (Vector A,Vector B)
{
    return Vector(A.x+B.x,A.y+B.y);
}
Vector operator - (Vector A,Vector B)
{
    return Vector(A.x-B.x,A.y-B.y);
}
Vector operator * (Vector A,double p)
{
    return Vector(A.x*p,A.y*p);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
bool operator == (const Point& a,const Point& b)//两点相等
{
    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
double Dot(Vector A,Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
    return sqrt(Dot(A,A));
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
Vector Normal(Vector A)//向量的单位法线，即左转90度
{
    double L=Length(A);
    return Vector(-A.y/L,A.x/L);
}

bool OnLeft(DLine L,Point p)//判断点p是否在有向直线L的左边
{
    return Cross(L.v,p-L.P)>0;
}
Point GetIntersection(DLine a,DLine b)//有向直线交点，假设交点唯一
{
    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(DLine* L,int n,Point* poly)//半平面交
{
    sort(L,L+n);
    int first,last;
    Point *p=new Point[n];
    DLine *q=new DLine[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;
}
Point poly[maxn];
DLine L[maxn];
int V[maxn],U[maxn],W[maxn];
int main()
{
    int n;
    while(~scanf("%d",&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 cnt=0,ok=1;
            double k=10000;
            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=Point(-c/a,0);
                    Vector v(b,-a);
                    L[cnt++]=DLine(p,v);
                }
            if(ok)
            {
                L[cnt++]=DLine(Point(0,0),Vector(0,-1));
                L[cnt++]=DLine(Point(0,0),Vector(1,0));
                L[cnt++]=DLine(Point(0,1),Vector(-1,1));
                if(!HalfPlaneIntersection(L,cnt,poly)) ok=0;
            }
            if(ok) printf("Yes
");
            else printf("No
");
        }
    }
    return 0;
}