Segments

Segments

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x₁y₁ x₂ y₂ follow, in which (x₁, y₁) and (x₂, y₂) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output "Yes!", if a line with desired property exists and must output "No!" otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10^-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
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

Sample Output

Yes!
Yes!
No!

题目的意思是，求一条直线，将已知线段投影到这条直线上有一个共同交点，判断是否存在这条直线。

这题要直接去求会很麻烦。但是想一想，将所有线段投影后，如果有共同点，则，这个点实际上是由n个点叠在一起的。如果我们把它展开了，以另一个视角观察将会是这样的：

所有线段在直线ansL上都会有一个共同的投影点，A。再观察，发现那些投影到A点的点都被直线L所经过。所以，题目就变成了，判断是否存在一条直线，与所有线段相交。

在实际操作时，只需枚举2n个点中任意两个点，判断经过这两点的直线是否符合要求。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 ;
 int n;
 ],v[];
 point operator - (point P,point Q){point ret; ret.x=P.x-Q.x,ret.y=P.y-Q.y; return ret;}
 double cross(point P,point Q){return P.x*Q.y-Q.x*P.y;}
 bool jug(point P,point Q){
     ;
     ; i<n; i++) ;
     ;
 }
 int main(){
     int T;
     for (scanf("%d",&T); T; T--){
         scanf("%d",&n);
         ; i<n; i++) scanf("%lf%lf%lf%lf",&u[i].x,&u[i].y,&v[i].x,&v[i].y);
         ; ) flag=;
         ; i<n-; i++) if (!flag)
             ; j<n; j++) if (!flag)
             if(jug(u[i],u[j])||jug(u[i],v[j])||jug(v[i],u[j])||jug(v[i],v[j])) flag=true;
         printf("%s\n",flag?"Yes!":"No!");
     }
     ;
 }