POJ 3304 Segments[直线与线段相交]

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13514		Accepted: 4331

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!

Source

Amirkabir University of Technology Local Contest 2006

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题目大意：给出n条线段两个端点的坐标，问所有线段投影到一条直线上，如果这些所有投影至少相交于一点就输出Yes！，否则输出No！。

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暴力枚举线段的交点组成直线然后判相交就行了

判断线段直线相交：两个叉积不同 (此题要额外判断两个sgn后都为0的情况(会当作同号哦) 就是重合 也算相交)

注意：

1.两点之间距离<1e-8是重合的

2.有可能恰好与一条线段重合

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=;
const double eps=1e-;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-; c=getchar();}
    while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
    return x*f;
}
inline int sgn(double x){
    if(abs(x)<eps) return ;
    else return x<?-:;
}
struct Vector{
    double x,y;
    Vector(double a=,double b=):x(a),y(b){}
    bool operator <(const Vector &a)const{
        return x<a.x||(x==a.x&&y<a.y);
    }
    void print(){
        printf("%lf %lf\n",x,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,double b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==&&sgn(a.y-b.y)==;}
 
double Cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}
double DisPP(Point a,Point b){
    Point t=a-b;
    return sqrt(t.x*t.x+t.y*t.y);
}
struct Line{
    Point s,t;
    Line(){}
    Line(Point p,Point v):s(p),t(v){}
}a[N];
bool isLSI(Line l1,Line l2){
    //puts("isLSI");
    //l1.s.print();l1.t.print();
    //l2.s.print();l2.t.print();
    Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
    //printf("%d %d end\n",sgn(Cross(v,u)),sgn(Cross(v,w)));
    return sgn(Cross(v,u))!=sgn(Cross(v,w))||!sgn(Cross(v,u));
}
int n,m;
double x,y,x2,y2;
bool check(Line l){
    if(sgn(DisPP(l.s,l.t))==) return false;
    //printf("check\n");
    //l.s.print();l.t.print();
    for(int i=;i<=n;i++)
        if(!isLSI(l,a[i])) return false;
    return true;
}
 
void solve(){
    for(int i=;i<=n;i++)
        for(int j=;j<=n;j++)
            if(check(Line(a[i].s,a[j].s))||check(Line(a[i].s,a[j].t))
               ||check(Line(a[i].t,a[j].s))||check(Line(a[i].t,a[j].t)))
                {puts("Yes!");return;}
    puts("No!");
}
 
int main(int argc, const char * argv[]) {
    int T=read();
    while(T--){
        n=read();
        for(int i=;i<=n;i++){
            scanf("%lf%lf%lf%lf",&x,&y,&x2,&y2);
            a[i]=Line(Point(x,y),Point(x2,y2));
        }
        solve();
    }
 
    return ;
}