poj3304 Segments

Subject description:

vjudge

POJ

answer:

Computational Geometry, very much pit.

First, it is clear that there is a straight line through all segments namely $ Yes $.

Consider enumerate any two endpoints.

Note also enumerate the same on a segment!

Note that $ eps = 1e-8 $!

Note that two points do not coincide sentenced!

Code:

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 150;
const double eps = 1e-8;
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
    Point operator - (const Point&a)const{return Point(x-a.x,y-a.y);}
    double operator ^ (const Point&a)const{return x*a.y-y*a.x;}
};
typedef Point Vector;
struct Line
{
    Point p;
    Vector v;
    Line(){}
    Line(Point p,Vector v):p(p),v(v){}
};
int n;
Point s[N][2];
int dcmp(double x)
{
    if(fabs(x)<eps)return 0;
    return x>0?1:-1;
}
bool diff(Line l,Point a,Point b)
{
    return dcmp(l.v^(a-l.p))*dcmp(l.v^(b-l.p))<=0;
}
int check(Point a,Point b)
{
    if(!dcmp(a.x-b.x)&&!dcmp(a.y-b.y))return 0;
    Line l = Line(a,b-a);
    for(int i=1;i<=n;i++)
        if(!diff(l,s[i][0],s[i][1]))return 0;
    return 1;
}
void work()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf%lf%lf",&s[i][0].x,&s[i][0].y,&s[i][1].x,&s[i][1].y);
    int FG = 0;
    for(int i=1;!FG&&i<=n;i++)for(int j=i;!FG&&j<=n;j++)
        if(check(s[i][0],s[j][0])||check(s[i][0],s[j][1])||check(s[i][1],s[j][0])||check(s[i][1],s[j][1]))FG=1;
    puts(FG?"Yes!":"No!");
}
int T;
int main()
{
    scanf("%d",&T);
    while(T--)work();
    return 0;
}

View Code