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");

        }

    }

}

poj3335 半平面交模版