C++(clang++11) 解法, 执行用时: 6ms, 内存消耗: 380K, 提交时间: 2021-02-14 15:03:17

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=10;
const double eps=1e-6,inf=0x3f3f3f3f;
int n,m,ans;
struct point{
    double x,y;
    point operator-(point a){
        return (point){x-a.x,y-a.y};
    }
    double operator*(point a){
        return x*a.y-a.x*y;
    }
    bool operator==(point a)const{
        return (fabs(x-a.x)<eps&&fabs(y-a.y)<eps);
    }
};
point no=(point){inf,inf},single=(point){-inf,-inf};
struct line{
    point s,t;
}l[N];
point turn(point a,line l){
    double fg=(l.t-l.s)*(a-l.s);
    if(fg>eps){
        point res;
        if(fabs(l.s.y-l.t.y)<eps){
            res.x=a.x,res.y=l.s.y*2-a.y;
        }
        else if(fabs(l.s.x-l.t.x)<eps){
            res.x=l.s.x*2-a.x,res.y=a.y;
        }
        else{
            double k1=(l.s.y-l.t.y)/(l.s.x-l.t.x),b1=l.s.y-k1*l.s.x;
            double k2=-1/k1,b2=a.y-k2*a.x;
            double x=(b2-b1)/(k1-k2),y=k1*x+b1;
            res.x=x*2-a.x,res.y=y*2-a.y;
        }
        return res;
    }
    return no;
}
void dfs(point a,int now){
    if(now==0){
        if(a.x>eps&&a.x<100-eps&&a.y>eps&&a.y<100-eps){
            ans++;
        }
        return;
    }
    point b=turn(a,l[now]);
    if(b==no)return;
    else{
        dfs(a,now-1);
        dfs(b,now-1);
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf%lf%lf",&l[i].s.x,&l[i].s.y,&l[i].t.x,&l[i].t.y);
    }
    scanf("%d",&m);
    point q;
    for(int i=1;i<=m;i++){
        ans=0;
        scanf("%lf%lf",&q.x,&q.y);
        dfs(q,n);
        printf("%d\n",ans);
    }
    return 0;
}