poj 3304 Segments 线段与直线相交

Segments

Time Limit: 1000MS		Memory Limit: 65536K

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 integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x₁ 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 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!

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+,M=2e6+,inf=1e9+;
const LL INF=1e18+,mod=,MOD=;
const double eps=1e-,pi=(*atan(1.0));

int sgn(double x)
{
    if(fabs(x) < eps)return ;
    if(x < ) return -;
    return ;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    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;
    }
};
double Cross(Point p0,Point p1,Point p2) //p0p1 X p0p2
{
    return (p1-p0)^(p2-p0);
}
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
{
    return sgn(Cross(l2.s,l1.s,l1.e))*sgn(Cross(l2.e,l1.s,l1.e)) <= ;
}
Point a[N],b[N];
double dist(Point a,Point b)
{
    return sqrt( (b - a)*(b - a) );
}
int check1(Line x,int n)
{
    if(sgn(dist(x.s,x.e))==)return ;
    for(int k=;k<=n;k++)
    {
        Line now=Line(a[k],b[k]);
        if(!Seg_inter_line(x,now))return ;
    }
    return ;
}
int check(int n)
{
    for(int i=;i<=n;i++)
    {
        for(int j=;j<=n;j++)
        {
            Line l1=Line(a[i],a[j]),l2=Line(a[i],b[j]),l3=Line(b[i],a[j]),l4=Line(b[i],b[j]);
            if(check1(l1,n)||check1(l2,n)||check1(l3,n)||check1(l4,n))return ;
        }
    }
    return ;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=;i<=n;i++)
        {
            double x1,y1,x2,y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            a[i]=Point(x1,y1),b[i]=Point(x2,y2);
        }
        if(check(n))printf("Yes!\n");
        else printf("No!\n");
    }
    return ;
}