POJ3304:Segments （几何：求一条直线与已知线段都有交点）

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!

题意：

已知N条线段，问是否存在直线X，使得这些线段在直线上的投影有公共点。

思路：

转化一下，就是问是是否存在直线X，使得X的垂线L与每个线段都有公共点。如果存在L，那么把其中一些满足条件的L平移，

可以使得它结果两个已知线段的端点。

注意一下：

需要判断是否两点重合，开始我用的方法是fabs(x1-x2)<eps，fabs(y1-y2)<eps则重合，改为dist<=eps则重合就AC了。

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
const double eps=1e-;
int N;
struct Cpoint
{
    double x,y;
    Cpoint(){}
    Cpoint(double xx,double yy):x(xx),y(yy){}
};
struct Cline
{
    Cpoint a,b;
    Cline(){}
    Cline(Cpoint aa,Cpoint bb):a(aa),b(bb){}
};
struct Vector
{
    double x,y;
    Vector(){}
    Vector(double xx,double yy):x(xx),y(yy){}
    double friend operator ^(Vector a,Vector b){
        return a.x*b.y-b.x*a.y;
    }
};
double dist(Cpoint a,Cpoint b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
Cline L[maxn];
bool check(Cpoint w, Cpoint v)
{
    if(dist(w,v)<=eps) return false;// 判断两点重合，开始用点直接比较WA了。
    Vector base(w.x-v.x,w.y-v.y);
    for(int i=;i<=N;i++){
        Vector e(L[i].a.x-v.x,L[i].a.y-v.y);
        Vector p(L[i].b.x-v.x,L[i].b.y-v.y);
        if((base^e)*(base^p)>eps) return false;
    } return true;
}
bool find()
{
    for(int i=;i<=N;i++)
      for(int j=;j<=N;j++){
         if(check(L[i].a,L[j].a)) return true;
         if(check(L[i].a,L[j].b)) return true;
         if(check(L[i].b,L[j].a)) return true;
         if(check(L[i].b,L[j].b)) return true;
    }
    return false;
}
int main()
{
    int T; double x,y,xx,yy;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        for(int i=;i<=N;i++){
            scanf("%lf%lf%lf%lf",&x,&y,&xx,&yy);
            L[i]=Cline(Cpoint(x,y),Cpoint(xx,yy));
        }
        if(find()) printf("Yes!\n");
        else printf("No!\n");
    }
    return ;
}