POJ - 3304 Segments (simple geometric)

Topic links: Click here

Subject to the effect: Given n line segments, now ask whether there is a straight line, so that all segments of this line there is the projection of a common point of intersection

Topic Analysis: The meaning of the questions a little abstract, need to change it, because there is a projection of a straight line to the intersection of all the segments, then on that straight line, starting from the intersection position, do another line in the direction of a line along the vertical, this line will be found and the n line segments exists a point of intersection, i.e. intersect, this subject was converted to whether there is a straight line, so that line segments have an intersection with the n

Because of the straight line is composed of two points, we enumerate all points on the line segment, as two points on a straight line configuration, each time to determine what the condition is satisfied, the time complexity is (2 * n) * (2 * n) * n, because n is relatively small, so the direct implementation on the line

There is a need to look at the details, there may be a case of coincidence of two points, this time about the need for special sentence

Code:

#include<iostream>
#include<cstdio> 
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<sstream>
using namespace std;
 
typedef long long LL;
 
const int inf=0x3f3f3f3f;

const int N=110;

const double eps = 1e-8;

int n;

int sgn(double x){
	if(fabs(x) < eps)return 0;
	if(x < 0)return -1;
	else return 1;
}

struct Point{
	double x,y;
	Point(){}
	Point(double _x,double _y){
		x = _x;
		y = _y;
	}
	void input(){
		scanf("%lf%lf",&x,&y);
	}
	double distance(Point p){
		return hypot(x-p.x,y-p.y);
	}
	Point operator -(const Point &b)const{
		return Point(x-b.x,y-b.y);
	}
	//叉积
	double operator ^(const Point &b)const{
		return x*b.y - y*b.x;
	}
	//点积
	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;
	}
	void input(){
		s.input();
		e.input();
	}
	//`直线和线段相交判断`
	//`-*this line   -v seg`
	//`2 规范相交`
	//`1 非规范相交`
	//`0 不相交`
	int linecrossseg(Line v){
		int d1 = sgn((e-s)^(v.s-s));
		int d2 = sgn((e-s)^(v.e-s));
		if((d1^d2)==-2) return 2;
		return (d1==0||d2==0);
	}
}line[N];

bool check(Line l)
{
	if(sgn(l.s.distance(l.e))==0)
		return false;
	for(int i=1;i<=n;i++)
		if(!l.linecrossseg(line[i]))
			return false;
	return true;
}

int main()
{
//	freopen("input.txt","r",stdin);
//	ios::sync_with_stdio(false);
	int w;
	cin>>w;
	while(w--)
	{
		vector<Point>point;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			line[i].input();
			point.push_back(line[i].s);
			point.push_back(line[i].e);
		}
		for(int i=0;i<point.size();i++)
			for(int j=0;j<point.size();j++)
				if(check(Line(point[i],point[j])))
				{
					puts("Yes!");
					goto end;
				}
		puts("No!");	
		end:;
	}
 
 
 
 
 
 
 
	
	
	
	
	
	
	
	
	
	return 0;
}