判斷線段和直線相交 POJ 3304

// 判斷線段和直線相交 POJ 3304
// 思路：
// 如果存在一條直線和所有線段相交，那么平移該直線一定可以經過線段上任意兩個點，并且和所有線段相交。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
const LL inf = 1e18;
const int N = 5000;
const double eps = 1e-8;

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

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.x+y*b.y;
    }
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    } 
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s,e=_e;
    }
};

double xmult(Point p0,Point p1,Point p2){
    return (p1-p0)^(p2-p0);
}

bool Seg_inter_line(Line l1,Line l2){
    return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e))<=0;
}

double dist(Point a,Point b){
    return sqrt((a-b)*(a-b));
}
Line line[N];
bool work(Line l1,int n){
    if(sgn(dist(l1.s,l1.e))==0) return false;
    for(int i=0;i<n;i++){
        if(Seg_inter_line(l1,line[i])==false) return false;
    }
    return true;
}
int main(){
    int n,T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        double x1,y1,x2,y2;
        for(int i=0;i<n;i++){
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i]=Line(Point(x1,y1),Point(x2,y2));
        }
        bool flag=false;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(work(Line(line[i].s,line[j].e),n)||work(Line(line[i].s,line[j].s),n)||work(Line(line[i].e,line[j].e),n)||work(Line(line[i].e,line[j].s),n)){
                    flag=true;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) puts("Yes!");
        else puts("No!");
    }
    return 0;
}

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