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-;
 const  db pi = acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 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)==&&cmp(y,k1.y)==;}
 };
 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[];
 };
 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}
 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))==;
 }
 int t,n;
 Line l[];
 bool slove(point s,point t){
     if(sign(s.dis(t)==))return false;
     for(int i=;i<=n;i++){
         if(sign(cross(s-l[i].p[],t-l[i].p[])*sign(cross(s-l[i].p[],t-l[i].p[])))>)
             return false;
     }
     return true;
 }
 int main(){
     scanf("%d",&t);
     while (t--){
         scanf("%d",&n);
         for(int i=;i<=n;i++){
             scanf("%lf%lf%lf%lf",&l[i].p[].x,&l[i].p[].y,&l[i].p[].x,&l[i].p[].y);
         }
         bool f=;
         for(int i=;i<=n;i++){
             for(int j=;j<=n;j++){
                 if(slove(l[i].p[],l[j].p[])){
                     f=;break;
                 }
                 else if(slove(l[i].p[],l[j].p[])){
                     f=;break;
                 }
                 else if(slove(l[i].p[],l[j].p[])){
                     f=;break;
                 }
                 else if(slove(l[i].p[],l[j].p[])){
                     f=;break; }
             }if(f)break;
         }
         if(!f)
             printf("No!\n");
         else
             printf("Yes!\n");
     }
 }
 /**
 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
  */