Segments POJ3304（计算几何+叉乘的应用）

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本文介绍了一种算法，用于判断给定的二维空间线段集合中是否存在一条直线，使得所有线段投影到该直线上至少有一点重合。通过列举线段端点组合，利用向量运算来验证这一条件。

SegmentsPOJ - 3304

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!

分析：

1.如果有一条直线通过所有的线段，那么这条直线的垂线就是答案

2.在所有线段端点所确定的直线里面，一定存在答案，如果不存在，那么就输出“No！”

关于上面两点简单证明一下：

1.垂足就满足题目条件

2.假设有一条通过所有线段的直线，那么小心的平移之，总可以与一个线段的端点重合，那么以这个端点为中心，小心的旋转直线，总可以正好过另一个端点，可能是这条线段的，也很可能是其他线段的，那么这个时候，这条仍然过所有线段的直线就过两个端点了

枚举思路：

先枚举同一条线段上的两个端点，在枚举两条线段上的端点的组合，而且注意如果两个端点是重点（横纵坐标都小于1e-8），那么就放掉这种情况，因为一个点是没办法确定一条直线

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps=1e-8;
struct Vector{
	double x,y;
	Vector(){}
	Vector(double x,double y):x(x),y(y){}
};
typedef Vector point;
Vector operator + (Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);};
Vector operator - (Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);};
Vector operator * (Vector a,Vector b){return Vector(a.x*b.x,a.y*b.y);};
Vector operator / (Vector a,Vector b){return Vector(a.x/b.x,a.y/b.y);};
double dot(Vector a,Vector b){
	return a.x*b.x+a.y*b.y;
}
double cross(Vector a,Vector b){
	return a.x*b.y-a.y*b.x;
}
struct segment{
	point p1,p2;
	segment(){}
	segment(point p1,point p2):p1(p1),p2(p2){} 
}ss[110];
int main(){
	int t;cin>>t;
	while(t--){
		int n;cin>>n;
		for(int i=0;i<n;i++){
			cin>>ss[i].p1.x>>ss[i].p1.y>>ss[i].p2.x>>ss[i].p2.y;
		}
		int flag=0;
		for(int i=0;i<n;i++){
			if(abs(ss[i].p1.x-ss[i].p2.x)<eps&&abs(ss[i].p1.y-ss[i].p2.y)<eps)continue;
			flag=1;
			Vector v=ss[i].p1-ss[i].p2;
			for(int j=0;j<n;j++){
				if(j==i)continue;
				Vector v1=ss[j].p1-ss[i].p2,v2=ss[j].p2-ss[i].p2;
				if(cross(v1,v)*cross(v2,v)>eps){
					flag=0;
					break;
				}
			} 
		}
		if(flag){
			cout<<"Yes!"<<endl;
			continue;
		}
		for(int i=0;i<n;i++){
			for(int j=i+1;j<n;j++){
				point p11=ss[i].p1,p12=ss[i].p2,p21=ss[j].p1,p22=ss[j].p2;
				if(abs(p11.x-p21.x)>eps&&abs(p11.y-p21.y)>eps){
					Vector v=p11-p21;
					flag=1;
					for(int k=0;k<n;k++){
						if(k==i||k==j)continue;
						Vector v1=ss[k].p1-p21,v2=ss[k].p2-p21;
						if(cross(v1,v)*cross(v2,v)>eps){
							flag=0;
							break;
						}
					}
					if(flag)break;
				}
				if(abs(p11.x-p22.x)>eps&&abs(p11.y-p22.y)>eps){
					Vector v=p11-p22;
					flag=1;
					for(int k=0;k<n;k++){
						if(k==i||k==j)continue;
						Vector v1=ss[k].p1-p22,v2=ss[k].p2-p22;
						if(cross(v1,v)*cross(v2,v)>eps){
							flag=0;
							break;
						}
					}
					if(flag)break;
				}
				if(abs(p12.x-p21.x)>eps&&abs(p12.y-p21.y)>eps){
					Vector v=p12-p21;
					flag=1;
					for(int k=0;k<n;k++){
						if(k==i||k==j)continue;
						Vector v1=ss[k].p1-p21,v2=ss[k].p2-p21;
						if(cross(v1,v)*cross(v2,v)>eps){
							flag=0;
							break;
						}
					}
					if(flag)break;
				}
				if(abs(p12.x-p22.x)>eps&&abs(p12.y-p22.y)>eps){
					Vector v=p12-p22;
					flag=1;
					for(int k=0;k<n;k++){
						if(k==i||k==j)continue;
						Vector v1=ss[k].p1-p22,v2=ss[k].p2-p22;
						if(cross(v1,v)*cross(v2,v)>eps){
							flag=0;
							break;
						}
					}
					if(flag)break;
				}
			}
			if(flag)break;
		}
		if(flag){
			cout<<"Yes!"<<endl;
			continue;
		}
		cout<<"No!"<<endl;
	}
	return 0;
}