[hdu3685]Rotational Painting 凸包 重心

大致题意：

　　给出一个多边形，问你有多少种放法可以使得多边形稳定得立在平面上。

　　先对多边形求重心，在求凸包，枚举凸包的边，如果重心没有在边的范围内，则不行

　　判断是否在范围内可用点积来判断

　　　

#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 n,m;
Point p[100005];
Point ch[100005];
Point cp;
void solve(){
    scanf("%d",&n);
    For(i,0,n-1) p[i].read();
    double tar=0,ar;
    cp.x=0;cp.y=0;
    For(i,2,n-1){
        ar=Cross(p[i]-p[0],p[i-1]-p[0])/2;
        tar+=ar;
        cp=cp+(p[i]+p[i-1]+p[0])*ar;
    }
    cp=cp/tar/3;
    sort(p,p+n);
    m=ConvexHull(p,n,ch);
    int ans=0;
    For(i,0,m-1) {
        int nxt=(i+1)%m;
        if(dcmp(Dot(ch[i]-ch[nxt],cp-ch[nxt]))>0 && dcmp(Dot(ch[nxt]-ch[i],cp-ch[i]))>0) ans++;
    }
    printf("%d
",ans);
}
int main(){
//    fre("in.txt","r",stdin);
    int t=0;
    cin>>t;
    For(i,1,t) solve();
    return 0;
}