poj 3304

我老人家要开始玩几何了！

。这个题有点自闭。

就是问是否存在一条直线经过所有了n条线段，(有交点).

我老人家愚昧不可救药，想了想决定先求出来 这两条直线的交点，然后看是否在线段上。但是一直写不对。。。

看了看题解发现可以直接用叉积，显然如果没有交点，那么线段在直线的一边，所以叉积就是正的，否则小于等于0.

#include <iostream>
#include <cmath>
#include <cstdio>
typedef double db;
const db eps=1e-8;
const  db pi = acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
struct point{
    db x,y;
    db abs(){return sqrt(x*x+y*y);}
    db dis(point k1){return ((*this)-k1).abs();}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator *(db k1)const {return (point){x*k1,y*k1};}
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
};
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
struct Line{
    point p[2];
};
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}
int inmid(point k1,point k2,point k3){//k3在[k1,k2]
    return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);
}
point getLL (point k1,point k2,point k3,point k4){//两直线交点
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
    return (k1*w2+k2*w1)/(w1+w2);
}
bool onS(point k1,point k2,point q){//q在[k1,k2]
    return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;
}
int t,n;
Line l[12306];
bool slove(point s,point t){
    if(sign(s.dis(t)==0))return false;
    for(int i=1;i<=n;i++){
        if(sign(cross(s-l[i].p[0],t-l[i].p[0])*sign(cross(s-l[i].p[1],t-l[i].p[1])))>0)
            return false;
    }
    return true;
}
int main(){
    scanf("%d",&t);
    while (t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lf%lf%lf%lf",&l[i].p[0].x,&l[i].p[0].y,&l[i].p[1].x,&l[i].p[1].y);
        }
        bool f=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(slove(l[i].p[0],l[j].p[0])){
                    f=1;break;
                }
                else if(slove(l[i].p[0],l[j].p[1])){
                    f=1;break;
                }
                else if(slove(l[i].p[1],l[j].p[0])){
                    f=1;break;
                }
                else if(slove(l[i].p[1],l[j].p[1])){
                    f=1;break; }
            }if(f)break;
        }
        if(!f)
            printf("No!
");
        else
            printf("Yes!
");
    }
}
/**
1
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0
 */