poj 3304

 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstdio>
 4 typedef double db;
 5 const db eps=1e-8;
 6 const  db pi = acos(-1);
 7 int sign(db k){
 8     if (k>eps) return 1; else if (k<-eps) return -1; return 0;
 9 }
10 int cmp(db k1,db k2){return sign(k1-k2);}
11 struct point{
12     db x,y;
13     db abs(){return sqrt(x*x+y*y);}
14     db dis(point k1){return ((*this)-k1).abs();}
15     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
16     point operator *(db k1)const {return (point){x*k1,y*k1};}
17     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
18     point operator / (db k1) const{return (point){x/k1,y/k1};}
19     int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
20 };
21 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
22 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
23 struct Line{
24     point p[2];
25 };
26 int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}
27 int inmid(point k1,point k2,point k3){//k3在[k1,k2]
28     return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);
29 }
30 point getLL (point k1,point k2,point k3,point k4){//两直线交点
31     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
32     return (k1*w2+k2*w1)/(w1+w2);
33 }
34 bool onS(point k1,point k2,point q){//q在[k1,k2]
35     return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;
36 }
37 int t,n;
38 Line l[12306];
39 bool slove(point s,point t){
40     if(sign(s.dis(t)==0))return false;
41     for(int i=1;i<=n;i++){
42         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)
43             return false;
44     }
45     return true;
46 }
47 int main(){
48     scanf("%d",&t);
49     while (t--){
50         scanf("%d",&n);
51         for(int i=1;i<=n;i++){
52             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);
53         }
54         bool f=0;
55         for(int i=1;i<=n;i++){
56             for(int j=1;j<=n;j++){
57                 if(slove(l[i].p[0],l[j].p[0])){
58                     f=1;break;
59                 }
60                 else if(slove(l[i].p[0],l[j].p[1])){
61                     f=1;break;
62                 }
63                 else if(slove(l[i].p[1],l[j].p[0])){
64                     f=1;break;
65                 }
66                 else if(slove(l[i].p[1],l[j].p[1])){
67                     f=1;break; }
68             }if(f)break;
69         }
70         if(!f)
71             printf("No!\n");
72         else
73             printf("Yes!\n");
74     }
75 }
76 /**
77 1
78 3
79 0.0 0.0 0.0 1.0
80 0.0 2.0 0.0 3.0
81 1.0 1.0 2.0 1.0
82  */