Codeforces Round #113 (Div. 2) B. Polygons Andrew求凸包-优快云博客

本文介绍了一种算法，用于判断一个多边形B的所有顶点是否严格位于另一个凸多边形A内部。通过合并两个多边形的顶点并计算合并后的凸包来实现这一目标。

You've got another geometrical task. You are given two non-degenerate polygons A and B as vertex coordinates. Polygon A is strictly convex. Polygon B is an arbitrary polygon without any self-intersections and self-touches. The vertices of both polygons are given in the clockwise order. For each polygon no three consecutively following vertices are located on the same straight line.

Your task is to check whether polygon B is positioned strictly inside polygon A. It means that any point of polygon B should be strictly inside polygon A. "Strictly" means that the vertex of polygon B cannot lie on the side of the polygon A.

Input

The first line contains the only integer n (3 ≤ n ≤ 105) — the number of vertices of polygon A. Then n lines contain pairs of integers xi, yi (|xi|, |yi| ≤ 109) — coordinates of the i-th vertex of polygon A. The vertices are given in the clockwise order.

The next line contains a single integer m (3 ≤ m ≤ 2·104) — the number of vertices of polygon B. Then following m lines contain pairs of integers xj, yj (|xj|, |yj| ≤ 109) — the coordinates of the j-th vertex of polygon B. The vertices are given in the clockwise order.

The coordinates of the polygon's vertices are separated by a single space. It is guaranteed that polygons A and B are non-degenerate, that polygon A is strictly convex, that polygon B has no self-intersections and self-touches and also for each polygon no three consecutively following vertices are located on the same straight line.

Output

Print on the only line the answer to the problem — if polygon B is strictly inside polygon A, print "YES", otherwise print "NO" (without the quotes).

Examples

input

output

YES

input

output

NO

input

output

NO

题意：给你两个多边形A，B，已知A为凸多边形，问B是否全部在A内（严格）；

思路：将所有点合并，查找凸包，求是否B上的点都不在凸包上；

　　　因为有一个严格的要求，对于扫描法的凸包，不一定可以找到相同斜率的点；

　　　对于Andrew求凸包，只删除一些点，使得形成凸包；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long

#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=1e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;
const double eps=(1e-8),pi=(4*atan(1.0));

int sgn(double x)
{
    if(fabs(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);
    }
    //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);
    }
    bool operator <(const Point p)const
    {
        if(x!=p.x)
        return x<p.x;
        return y<p.y;
    }
};
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)) == 0)
        {
            if(sgn((s-b.e)^(b.s-b.e)) == 0) return make_pair(0,res);//重合
            else return make_pair(1,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(2,res);
    }
};

double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
const int MAXN = 200010;
Point listt[MAXN];
int Stack[MAXN],top;

int convexhull(int n) /*建立凸包*/
{
    sort(listt,listt+n);
    int m=0;
    for(int i=0; i<n; i++) {
        while(m>1 && sgn((listt[Stack[m-1]]-listt[Stack[m-2]])^(listt[i]-listt[Stack[m-2]]))<0)
            m--;
        Stack[m++]=i;
    }
    int k=m;
    for(int i=n-2; i>=0; i--) {
        while(m>k && sgn((listt[Stack[m-1]]-listt[Stack[m-2]])^(listt[i]-listt[Stack[m-2]]))<0)
            m--;
        Stack[m++]=i;
    }
    if(n>1) m--;
    return m;
}
set<Point>s;
int n,m;
int FIND()
{
    //for(int i=0;i<top;i++)
        //cout<<listt[Stack[i]].x<<" "<<listt[Stack[i]].y<<endl;
    for(int i=0;i<top;i++)
        if(s.find(listt[Stack[i]])!=s.end())return 1;
    return 0;
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        double x,y;
        scanf("%lf%lf",&x,&y);
        listt[i]=Point(x,y);
    }
    scanf("%d",&m);
    for(int i=0;i<m;i++)
    {
        double x,y;
        scanf("%lf%lf",&x,&y);
        listt[i+n]=Point(x,y);
        s.insert(listt[i+n]);
    }
    top=convexhull(n+m);if(FIND())printf("NO\n");
    else printf("YES\n");
    return 0;
}