Segments

Description

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 x₁ y₁ 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!

Source

Amirkabir University of Technology Local Contest 2006    
题目大意：给出n条线段两个端点的坐标，问所有线段投影到一条直线上，如果这些所有投影至少相交于一点就输出Yes！，否则输出No！。
解题思路：如果有存在这样的直线，过投影相交区域作直线的垂线，该垂线必定与每条线段相交，问题转化为问是否存在一条线和所有线段相交   直线肯定经过两个端点。 枚举端点，判断直线和线段是否相交。   细节要注意，判断重合点。   还有就是加入只有一条线段的话，刚好直线是过同一条直线的。 所以保险的做法是枚举所有的两个端点，包括同一条直线的。  

/************************************************************
 * Author        : kuangbin
 * Email         : kuangbin2009@126.com 
 * Last modified : 2013-07-13 20:57
 * Filename      : POJ3304Segments.cpp
 * Description   : 
 * *********************************************************/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
};
double xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2
{
    return (p1-p0)^(p2-p0);
}
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
{
    return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0;
}
double dist(Point a,Point b)
{
    return sqrt( (b - a)*(b - a) );
}
const int MAXN = 110;
Line line[MAXN];
bool check(Line l1,int n)
{
    if(sgn(dist(l1.s,l1.e)) == 0 )return false;
    for(int i = 0;i < n;i++)
        if(Seg_inter_line(l1,line[i]) == false)
            return false;
    return true;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        double x1,y1,x2,y2;
        for(int i = 0; i < n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i] = Line(Point(x1,y1),Point(x2,y2));
        }
        bool flag = false;
        for(int i = 0;i < n;i++)
            for(int j = 0; j < n;j++)
                if(check(Line(line[i].s,line[j].s),n) || check(Line(line[i].s,line[j].e),n)
                        || check(Line(line[i].e,line[j].s),n) || check(Line(line[i].e,line[j].e),n) )
                {
                    flag = true;
                    break;
                }
        if(flag)
            printf("Yes!\n");
        else printf("No!\n");
    }
    return 0;
}