模板题
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define root 1,n,1
#define mid ((l+r)>>1)
#define ll long long
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****\n");
using namespace std;
const int MAXN=+;
int sum[MAXN<<],lsum[MAXN<<],rsum[MAXN<<];
int n,m,tt;
const double eps = 1e-;
const double PI = acos(-1.0);
int sgn(double x)
{
if(fabs(x) < eps)return ;
if(x < )return -;
else return ;
}
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);
}
//叉积
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
//绕原点旋转角度B（弧度值），后x,y的变化
void transXY(double B)
{
double tx = x,ty = y;
x = tx*cos(B) - ty*sin(B);
y = tx*sin(B) + ty*cos(B);
}
};
//*判断点在线段上
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
//只有第一个值为2时，交点才有意义
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == )
{
if(sgn((s-b.e)^(b.s-b.e)) == )
return make_pair(,res);//重合
else return make_pair(,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(,res);
}
};
bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P)) == &&
sgn((P.x - L.s.x) * (P.x - L.e.x)) <= &&
sgn((P.y - L.s.y) * (P.y - L.e.y)) <= ;
}
bool inter(Line l1,Line l2)
{
return
max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= &&
sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= ;
}
//*判断点在任意多边形内
//射线法，poly[]的顶点数要大于等于3,点的编号0~n-1
//返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point p,Point poly[],int n)
{
int cnt;
Line ray,side;
cnt = ;
ray.s = p;
ray.e.y = p.y;
ray.e.x = -100000000000.0;//-INF,注意取值防止越界
for(int i = ;i < n;i++)
{
side.s = poly[i];
side.e = poly[(i+)%n];
if(OnSeg(p,side))return ;
//如果平行轴则不考虑
if(sgn(side.s.y - side.e.y) == )
continue;
if(OnSeg(side.s,ray))
{
if(sgn(side.s.y - side.e.y) > )cnt++;
}
else if(OnSeg(side.e,ray))
{
if(sgn(side.e.y - side.s.y) > )cnt++;
}
else if(inter(ray,side))
cnt++;
}
if(cnt % == )return ;
else return -;
} Point a[MAXN],b;
int main()
{
int i,j,k;
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
#endif
while(scanf("%d",&n)!=EOF)
{
for(i=;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
}
scanf("%d",&m);
for(i=;i<m;i++)
{
scanf("%lf%lf",&b.x,&b.y);
if(inPoly(b,a,n)!=-)
{
printf("Yes\n");
}
else printf("No\n");
}
}
}