LA2218 Triathlon

题意

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分析

设出长度\(x,y,1-x-y\)，就是关于它们的二元一次不等式，判断有没有解。

可以用半平面交来解决。

x/V[i]+y/U[i]+(1-x-y)/W[i] < x/V[j]+y/U[j]+(1-x-y)/W[j]
ax+by+c>0

然后关于这些不等式似乎只能先化为我所熟悉的\(y=kx+b\)然后再做？我没有想出更好的解决方法。

枚举每一个点再枚举点集，时间复杂度\(O(T n^2 \log n)\)

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
    rg T data=0;
    rg int w=1;
    rg char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data*w;
}
template<class T>T read(T&x)
{
    return x=read<T>();
}
using namespace std;
typedef long long ll;

struct Point
{
    double x,y;
    
    Point(double x=0,double y=0)
    :x(x),y(y){}
};
typedef Point Vector;

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

Vector operator-(co Vector&A,co Vector&B)
{
    return Vector(A.x-B.x,A.y-B.y);
}

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

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

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

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

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

double PolygonArea(vector<Point>p)
{
    int n=p.size();
    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;
}

struct Line
{
    Point P;
    Vector v;
    
    Line(Point P=0,Vector v=0)
    :P(P),v(v){}
    
    double angle()co
    {
        return atan2(v.y,v.x);
    }
    
    bool operator<(co Line&rhs)co
    {
        return angle()<rhs.angle();
    }
};

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

Point LineLineIntersection(co Line&a,co 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;
}

co double eps=1e-6;

vector<Point>HalfplaneIntersection(vector<Line>L)
{
    int n=L.size();
    sort(L.begin(),L.end());
    
    int first,last;
    vector<Point>p(n);
    vector<Line>q(n);
    vector<Point>ans;
    
    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]=LineLineIntersection(q[last-1],q[last]);
    }
    while(first<last&&!OnLeft(q[first],p[last-1]))
        --last;
    if(last-first<=1)
        return ans;
    p[last]=LineLineIntersection(q[last],q[first]);
    
    for(int i=first;i<=last;++i)
        ans.push_back(p[i]);
    return ans;
}

co int N=100;
int V[N],U[N],W[N];

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;++i)
        {
            read(V[i]);read(U[i]);read(W[i]);
        }
        for(int i=0;i<n;++i)
        {
            bool ok=1;
            double k=1e4;
            vector<Line>L;
            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.push_back(Line(P,v));
                }
            if(ok)
            {
                L.push_back(Line(Point(0,0),Vector(0,-1)));
                L.push_back(Line(Point(0,0),Vector(1,0)));
                L.push_back(Line(Point(0,1),Vector(-1,1)));
                vector<Point>poly=HalfplaneIntersection(L);
                if(poly.empty())
                    ok=0;
            }
            puts(ok?"Yes":"No");
        }
    }
    return 0;
}