[HDU - 5408] CRB and Farm

　　题意：

　　　　给出一个由n个点组成的凸包，以及凸包内的k个点，问是否能够在凸包上选择最多2k个点构造一个新的

　　凸包，使得新的凸包覆盖原来的k个点。

　　　　

　　　　要求2k个点覆盖原本的k个点，只要对原k个点构造凸包，然后选择该凸包内一点O与该凸包的顶点连一条射线，其

　　与大凸包相交的边的两端点即为要保留的点，小凸包的顶点个数<=k，所有结果最后一定存在。

　　

　　　　

 #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-7
 #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 1e9+9
 #define Vector Point
 const int Mod=1e9+;
 typedef unsigned long long ull;
 typedef long long ll;
 int dcmp(double x){
     if(fabs(x)<=eps) return ;
     return x<?-:;
 }
 double ox,oy;
 struct Point{
     double x,y,id;
     Point(){}
     Point(double x,double y):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); }
     bool operator == (const Point &a)const{ return dcmp(x-a.x)== && dcmp(y-a.y)==; }
 };
 double Dot(Vector a,Vector b) { return a.x*b.x+a.y*b.y; }
 ll dis(Vector a) { return sqrt(Dot(a,a)); }
 double Cross(Vector a,Vector b) { return a.x*b.y-a.y*b.x; }
 int ConvexHull(Point *p,int n,Point *ch){
     sort(p+,p++n);
     int m=;
     For(i,,n){
         while(m> && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=) m--;
         ch[m++]=p[i];
     }
     int k=m;
     Fore(i,n-,) {
         while(m>k && Cross(ch[m-]-ch[m-],p[i]-ch[m-])<=) m--;
         ch[m++]=p[i];
     }
     if(n>) m--;
     return m;
 }
 Point getInsect(Point p,Vector v,Point q,Vector w) {
     Vector u=p-q;
     double t=Cross(w,u)/Cross(v,w);
     return p+v*t;
 }
 Point p[],pp[],ch[],o;
 bool chk(Point p1,Point p2,Point q1,Point q2) {
     double c1=Cross(p2-p1,q1-p1),c2=Cross(p2-p1,q2-p1);
     double c3=Dot(p2-p1,q1-p1),c4=Dot(p2-p1,q2-p1);
     if(dcmp(c1)*dcmp(c2)<) {
         Point cp=getInsect(p1,p2-p1,q1,q2-q1);
         return Dot(cp-p1,p2-p1)>;
     }
     return ;
 }
 int n,k,res;
 set<int>ans;
 void solve(){
     ans.clear();
     scanf("%d",&n);
     For(i,,n-) p[i].read();
     scanf("%d",&k);
     For(i,,k) pp[i].read();
     int m=ConvexHull(pp,k,ch);
     o=ch[];
     For(i,,m-) o=(o+ch[i])/;
     res=;
     For(i,,m-) {
         while(!chk(o,ch[i],p[res],p[(res+)%n])) res=(res+)%n;
         ans.insert(res);ans.insert((res+)%n);
     }
     set<int>::iterator it;
     cout<<"Yes"<<endl;
     cout<<ans.sz<<endl;
     for(it=ans.begin();it!=ans.end();it++){
         if(it!=ans.begin()) printf(" ");
         printf("%d",(*it)+);
     }
     cout<<endl;
 }
 int main(){
 //    fre("in.txt","r",stdin);
     int t=;
     cin>>t;
     For(i,,t) solve();
     return ;
 }