但是 algo log ret cst ive follow amp dimens

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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!

題目的意思是，求一條直線，將已知線段投影到這條直線上有一個共同交點，判斷是否存在這條直線。

這題要直接去求會很麻煩。但是想一想，將所有線段投影後，如果有共同點，則，這個點實際上是由n個點疊在一起的。如果我們把它展開了，以另一個視角觀察將會是這樣的：

所有線段在直線ansL上都會有一個共同的投影點，A。再觀察，發現那些投影到A點的點都被直線L所經過。所以，題目就變成了，判斷是否存在一條直線，與所有線段相交。

在實際操作時，只需枚舉2n個點中任意兩個點，判斷經過這兩點的直線是否符合要求。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 const double eps=1e-8;
 7 int n;
 8 struct point{double x,y;}u[105],v[105];
 9 point operator - (point P,point Q){point ret; ret.x=P.x-Q.x,ret.y=P.y-Q.y; return ret;}
10 double cross(point P,point Q){return P.x*Q.y-Q.x*P.y;}
11 bool jug(point P,point Q){
12     if (fabs(P.x-Q.x)<eps&&fabs(P.y-P.y)<eps) return 0;
13     for (int i=0; i<n; i++) if (cross(P-u[i],Q-u[i])*cross(P-v[i],Q-v[i])>eps) return 0;
14     return 1;
15 }
16 int main(){
17     int T;
18     for (scanf("%d",&T); T; T--){
19         scanf("%d",&n);
20         for(int i=0; i<n; i++) scanf("%lf%lf%lf%lf",&u[i].x,&u[i].y,&v[i].x,&v[i].y);
21         bool flag=0; if (n<3) flag=1;
22         for(int i=0; i<n-1; i++) if (!flag)
23             for(int j=i+1; j<n; j++) if (!flag)
24             if(jug(u[i],u[j])||jug(u[i],v[j])||jug(v[i],u[j])||jug(v[i],v[j])) flag=true;
25         printf("%s\n",flag?"Yes!":"No!");
26     }
27     return 0;
28 }

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