[Codeforces166B]Polygons 凸包

大致题意：

　　　　给你一个凸多边形A，和一个任意多边形B，判断B是否在A的内部

　　先对A的点集建凸包，然后枚举B中的点，二分判断是否在A的内部。

　　二分时可用叉积判断，详细见代码

　　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++) 
#define Fore(i,a,b) for(int i=a;i>=b;i--) 
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e6+7
#define Vector Point
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
struct Point{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
    bool operator < (const Point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    Point operator - (const Point &a)const{
        return Point(x-a.x,y-a.y);
    }
    Point operator + (const Point &a)const{
        return Point(x+a.x,y+a.y);
    }
    Point operator * (const double &a)const{
        return Point(x*a,y*a);
    }
    Point operator / (const double &a)const{
        return Point(x/a,y/a);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    void out(){
        cout<<"debug: "<<x<<" "<<y<<endl;
    }
    bool operator == (const Point &a)const{
        return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
    }
};
double Dot(Vector a,Vector b) {
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a) {
    return sqrt(Dot(a,a));
}
double Cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
int ConvexHull(Point *p,int n,Point *ch){
    int m=0;
    For(i,0,n-1) {
        while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    Fore(i,n-2,0){
        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--;
    return m;
}
int n1,n2,m;
Point p1[100005],p2[100005];
Point ch[100005];
bool check(Point tp){
    if(Cross(tp-ch[0],ch[1]-ch[0])>=0 || Cross(tp-ch[0],ch[m-1]-ch[0])<=0) return 0;
    int l=0,r=m-1;
    int now=l;
    while(l<=r) {
        int mid=l+r>>1;
        if(Cross(ch[mid]-ch[0],tp-ch[0])>0) l=mid+1,now=mid;
        else r=mid-1;
    }
    return Cross(ch[(now+1)%m]-ch[now],tp-ch[now])>0;
}
void solve(){
    cin>>n1;
    For(i,0,n1-1) p1[i].read();
    cin>>n2;
    For(i,0,n2-1) p2[i].read();
    sort(p1,p1+n1);
    m=ConvexHull(p1,n1,ch);
    int mk=1;
    For(i,0,n2-1) {
        if(!mk) break;
        if(!check(p2[i])) mk=0;
    }
    if(mk) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}
int main(){
//    fre("in.txt","r",stdin);
    int t=0;
    solve();
    return 0;
}