POJ - 3304 ：Segments__判断直线和线段是否相交

最新推荐文章于 2020-07-19 22:59:12 发布

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最新推荐文章于 2020-07-19 22:59:12 发布

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分类专栏：计算几何文章标签： struct output ini each input

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本文介绍了一种算法，用于判断二维空间中是否存在一条直线能够与给定的所有线段相交。通过枚举线段端点并检查每一对端点确定的直线是否与所有线段有交点来实现。

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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 integern ≤ 100 showing the number of segments. After that, n lines containing four real numbersx₁ 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 numbersa 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!

Hint

Source

Amirkabir University of Technology Local Contest 2006

//题意：给出一些线段，问是否存在某条直线与所有线段都相交。

//分析：假设存在一条直线与所有线段都相交，那么这条直线可以通过旋转或者平移必定可以达到一个临界状态即恰好交与某两条线段的端点。所以只要枚举每两条线段的端点假定确定的一条直线（4种情况），判断这样假定的直线是否与所有线段相交即可！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 105 
#define eps 1e-8

struct point 
{
	double x,y;
};
struct line
{
	point sp,ep;
}L[maxn];
int n;

double Len_ab(point a,point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double x_multi(point p1,point p2,point p3)
{
	return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);
}

bool Cross(point p1,point p2,point p3,point p4) //判断直线(p1,p2)和线段(p3,p4)是否相交
{
	double a=x_multi(p1,p2,p3);
	double b=x_multi(p1,p2,p4);
	if(fabs(a)<=eps) return true;
	if(fabs(b)<=eps) return true;
	if(a*b<=eps) return true;
	return false;
}

bool OK(point a,point b)
{
	if(Len_ab(a,b)<eps) //去重点
		return false;
	int i;
	for(i=0;i<n;i++)
		if(!Cross(a,b,L[i].sp,L[i].ep))
			return false;
	return true;
}

int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%lf%lf%lf%lf",&L[i].sp.x,&L[i].sp.y,&L[i].ep.x,&L[i].ep.y);
		if(n==1)
		{
			puts("Yes!");
			continue;
		}
		for(i=0;i<n;i++)
			for(j=i+1;j<n;j++)
			{
				if(OK(L[i].sp,L[j].sp)||OK(L[i].sp,L[j].ep)||OK(L[i].ep,L[j].sp)||OK(L[i].ep,L[j].ep))
				{
					puts("Yes!");
					goto end;
				}
			}
		puts("No!");
		end:;
	}
	return 0;
}