poj3335 半平面交模版

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef double dd;
#define N 200
struct P{
       dd x,y;
       P(dd a=0,dd b=0){
       x=a,y=b;        
    }
       P operator+(P a){
       return P(x+a.x,y+a.y);
    }
    P operator-(P a){
        return P(x-a.x,y-a.y);
    }
    P operator*(dd a){
       return P(x*a,y*a);    
    }
    void out(){
        cout<<x<<" "<<y<<"   ";    
    }
}z[N];
struct ply{
   int siz;
   P a[N];    
   ply(){
     siz=0;        
   }
};
dd mul(P x,P y,P z){
    return (y.x-x.x)*(z.y-x.y)-(z.x-x.x)*(y.y-x.y);
}
P jiao(P a,P b,P c,P d){
        dd a1=mul(c,a,b),a2=mul(d,b,a);
        return c+(d-c)*(a1/(a1+a2));
}
int cas,n;
ply add(ply x,P a,P b){
    ply res;
    for(int i=0;i<x.siz;i++){
        P t1=x.a[i];
        P t2=x.a[(i+1)%x.siz];
        dd c=mul(a,t1,b);
        dd d=mul(a,t2,b);
        if(c>=0)res.a[res.siz++]=t1;
        if(c*d<0){
            res.a[res.siz++]=jiao(t1,t2,a,b);    
            }
    }
    return res;
}
bool ok(){
    ply now;
    now.siz=4;
    now.a[0]=P(-1e10,-1e10);
    now.a[1]=P(-1e10,1e10);
    now.a[2]=P(1e10,1e10);
    now.a[3]=P(1e10,-1e10);
    for(int i=0;i<n;i++){
        now=add(now,z[i],z[(i+1)%n]);
        if(!now.siz)return 0;
    }
    return 1;
}
int main(){
    scanf("%d",&cas);
    while(cas--){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&z[i].x,&z[i].y);
        }  
        if(ok()){
         printf("YES\n");    
        }else{
           printf("NO\n");    
        }        
    }    
}