POJ 3304 Segments

题意：判断是否存在一条直线与所有线段相交

思路：枚举线段端点

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstdio>
using namespace std;
#define eps 1e-8
struct point
{
    double x,y;
    point()
    {

    }
    point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
} p[500];
int judge(int a,int b)
{
    if(a%2==1) a--;
    if(a==b) return 1;
    return 0;
}
int banana(point a,point  b,point c,point d)
{
    double cd=(b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
    double dd=(b.x-a.x)*(d.y-a.y)-(d.x-a.x)*(b.y-a.y);
    if(fabs(cd)<eps) return 1;
    if(fabs(dd)<eps) return 1;
    if(cd*dd<=eps) return 1;
    return 0;
}
double dist(int a,int b)
{
    return hypot(fabs(p[a].x-p[b].x),fabs(p[a].y-p[b].y));
}
int main()
{
    int T,n,i,j,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        n*=2;
        for(i=0; i<n; i+=2)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            scanf("%lf%lf",&p[i+1].x,&p[i+1].y);
        }
        int flag=0;
        for(i=0; i<n; i++)
        {
            for(j=i+1; j<n; j++)
            {
                if(judge(j,i)) continue;
                if(dist(i,j)<eps) continue;
                flag=0;
                for(k=0; k<n; k+=2)
                {
                    if(judge(i,k)) continue;
                    if(judge(j,k)) continue;
                    if(banana(p[i],p[j],p[k],p[k+1])==0)
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag==0) break;
            }
            if(flag==0) break;
        }
        if(flag==0) printf("Yes!
");
        else printf("No!
");
    }
    return 0;
}