poj3304 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 x₁y₁x₂y₂ 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：问你是有存在一条线和n条线都有交点
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，
若存在一条直线与所有线段相机相交，此时该直线必定经过这些线段的某两个端点，所以枚举任意两个端点即可。
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const double eps=1e-8;
bool f;
int n;
struct point 
{
	double x,y;
};

struct vector
{
	point start,end;
};
point p;
vector line[1000];
double multi(point p1,point p2,point p0)
{
	return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool fun(point a,point b)
{
	if(fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps)
 return false;
	for(int i=0;i<n;i++)
	{
		if(multi(line[i].start,a,b)*multi(line[i].end,a,b)>eps)
		return false;
	}

	return true;
	
}

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		f=false;
		cin>>n;
		for(int i=0;i<n;i++)
		cin>>line[i].start.x>>line[i].start.y>>line[i].end.x>>line[i].end.y;
			if(n<3)
			f=true;
		for(int i=0;i<n&&!f;i++)
		{
		//	if(fun(line[i].end,line[i].start))
		//	f=true;
			for(int j=i+1;j<n&&!f;j++)
			if(fun(line[i].start,line[j].end)||
			fun(line[i].end,line[j].start)||
			fun(line[i].end,line[j].end)||
			fun(line[i].start,line[j].start))
			f=true;	
		}
		if(!f)
		printf("No!\n");
		else
		printf("Yes!\n");	
	}
	return 0;
 }