題意

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分析

就判斷兩個點集的凸包相離即可。

需要滿足如下兩個條件：

1. 兩凸包上任意兩條線段不相交。
2. 兩凸包上任意一點不在另一凸包的內部。

時間複雜度\(O(T n m)\)

程式碼

#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;

co double eps=1e-10;

double dcmp(double x)
{
    return fabs(x)<eps?0:(x<0?-1:1);
}

struct Point
{
    double x,y;
    
    Point(double x=0,double y=0)
    :x(x),y(y){}
    
    bool operator<(co Point&rhs)co
    {
        return x<rhs.x||(x==rhs.x&&y<rhs.y);
    }
    
    bool operator==(co Point&rhs)co
    {
        return x==rhs.x&&y==rhs.y;
    }
};
typedef Point Vector;

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

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;
}

bool SegmentProperIntersection(co Point&a1,co Point&a2,co Point&b1,co Point&b2)
{
    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
           c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(co Point&p,co Point&a1,co Point&a2)
{
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

vector<Point>ConvexHull(vector<Point>p)
{
    sort(p.begin(),p.end());
    p.erase(unique(p.begin(),p.end()),p.end());
    
    int n=p.size();
    int m=0;
    vector<Point>ch(n+1);
    for(int i=0;i<n;++i)
    {
        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
            --m;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;--i)
    {
        while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
            --m;
        ch[m++]=p[i];
    }
    if(n>1)
        m--;
    ch.resize(m);
    return ch;
}

int PointInPolygon(co Point&p,co vector<Point>&poly)
{
    int wn=0;
    int n=poly.size();
    for(int i=0;i<n;++i)
    {
        co Point&p1=poly[i];
        co Point&p2=poly[(i+1)%n];
        if(p1==p||p2==p||OnSegment(p,p1,p2))
            return -1;
        int k=dcmp(Cross(p2-p1,p-p1));
        int d1=dcmp(p1.y-p.y);
        int d2=dcmp(p2.y-p.y);
        if(k>0&&d1<=0&&d2>0)
            ++wn;
        if(k<0&&d2<=0&&d1>0)
            --wn;
    }
    if(wn!=0)
        return 1;
    return 0;
}

bool ConvexPolygonDisjoint(co vector<Point>ch1,co vector<Point>&ch2)
{
    int c1=ch1.size();
    int c2=ch2.size();
    for(int i=0;i<c1;++i)
        if(PointInPolygon(ch1[i],ch2)!=0)
            return 0;
    for(int i=0;i<c2;++i)
        if(PointInPolygon(ch2[i],ch1)!=0)
            return 0;
    for(int i=0;i<c1;++i)
        for(int j=0;j<c2;++j)
            if(SegmentProperIntersection(ch1[i],ch1[(i+1)%c1],ch2[j],ch2[(j+1)%c2]))
                return 0;
    return 1;
}

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    int n,m;
    while(read(n)|read(m))
    {
        vector<Point>P1,P2;
        for(int i=0;i<n;++i)
            P1.push_back(Point(read<int>(),read<int>()));
        for(int i=0;i<m;++i)
            P2.push_back(Point(read<int>(),read<int>()));
        if(ConvexPolygonDisjoint(ConvexHull(P1),ConvexHull(P2)))
            puts("Yes");
        else
            puts("No");
    }
    return 0;
}