凸包模板***

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define PF(x) cout << "debug: " << x << " ";
#define EL cout << endl;
#define PC(x) puts(x);
typedef long long ll;
#define CLR(x, v) sizeof (x, v, sizeof(x))
using namespace std;
const int INF = 0x5f5f5f5f;
const int maxn = 4e5 + 10;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int sgn(double x){
    if(abs(x) < eps) return 0;
    if(x < 0) return -1;
    else  return 1;
}
struct Point
{
       double x,y;
       Point(){}
       Point(double _x,double _y)
       {
                    x=_x;y=_y;
       }
       Point operator - (const Point &b) const
       {
             return Point(x-b.x,y-b.y);
       }
       bool operator < (const Point &b) const
       {
            return (x<b.x)||(x==b.x&&y<b.y);
       }
};
//*两点间距离
int n,m;
Point p[maxn],pb[maxn],res[maxn];
double dist(Point a,Point b)
{
    return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
//相对于p[0]的极角排序
int cmp(Point a,Point b)
{
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
double cross(Point a,Point b)
{
       return a.x*b.y-b.x*a.y;
}
int andrew(int n)//从左下开始逆时针凸包，res中存的既是按顺序的凸包点，返回的是0~m-1共m个点
{
    sort(p,p+n);
    int m=0;
    for (int i=0;i<n;i++)
    {
        while (m>1&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<0) --m;
        res[m++]=p[i];
    }
    int k=m;
    for (int i=n-2;i>=0;--i)
    {
        while (m>k&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<0) --m;
        res[m++]=p[i];
    }
    if (m>1) --m;
    return m;
}
int main(){
    //freopen("in.txt","r",stdin);
  cin>>n;
  for(int i = 0;i < n;i++)
    scanf("%lf%lf",&p[i].x,&p[i].y);
  cin>>m;
  for(int i = 0;i < m;i++){
    scanf("%lf%lf",&p[i+n].x,&p[i+n].y);
    pb[i] = p[i+n];
  }
  int num = andrew(n+m);
 // cout<<num<<endl;
  int fg = 0;
  sort(pb,pb+m);
  for(int i = 0;i < num;i++){
   // cout<<res[i].x<<" "<<res[i].y<<endl;
    int tmp = lower_bound(pb,pb+m,res[i]) - pb;
    if(sgn(dist(res[i],pb[tmp])) == 0){
        fg = 1;
      //  break;
    }
  }
    if(fg) cout<<"NO"<<endl;
    else cout<<"YES"<<endl;

    return 0;
}