hdu 1558 线段相交+并查集

题意：要求相交的线段都要塞进同一个集合里

sol：并查集+判断线段相交即可。n很小所以n^2就可以水过

 #include <iostream>
 #include <cmath>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 int f[];
 char ch;
 int tmp,n;
 double X1,X2,Y1,Y2;

    #define eps 1e-8
    #define PI acos(-1.0)//3.14159265358979323846
    //判断一个数是否为0,是则返回true,否则返回false
    #define zero(x)(((x)>0?(x):-(x))<eps)
    //返回一个数的符号，正数返回1，负数返回2，否则返回0
    #define _sign(x)((x)>eps?1:((x)<-eps?2:0))
   struct point
   {
       double x,y;
       point(){}
       point(double xx,double yy):x(xx),y(yy)
       {}
   };
   struct line
   {
       point a,b;
       line(){}      //默认构造函数
       line(point ax,point bx):a(ax),b(bx)
       {}
   }l[];//直线通过的两个点，而不是一般式的三个系数

    //求矢量[p0,p1],[p0,p2]的叉积
    //p0是顶点
    //若结果等于0，则这三点共线
    //若结果大于0，则p0p2在p0p1的逆时针方向
    //若结果小于0，则p0p2在p0p1的顺时针方向
    double xmult(point p1,point p2,point p0)
    {
        return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    }
    //计算dotproduct(P1-P0).(P2-P0)
    double dmult(point p1,point p2,point p0)
    {
        return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);
    }
    //两点距离
    double distance(point p1,point p2)
    {
        return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
    }
    //判三点共线
    int dots_inline(point p1,point p2,point p3)
    {
        return zero(xmult(p1,p2,p3));
    }
    //判点是否在线段上,包括端点
    int dot_online_in(point p,line l)
    {
        return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;
    }
    //判点是否在线段上,不包括端点
    int dot_online_ex(point p,line l)
    {
        return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y))&&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y));
    }
    //判两点在线段同侧,点在线段上返回0
    int same_side(point p1,point p2,line l)
    {
        return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;
    }
    //判两点在线段异侧,点在线段上返回0
    int opposite_side(point p1,point p2,line l)
    {
        return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)<-eps;
    }
    //判两直线平行
    int parallel(line u,line v)
    {
        return zero((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));
    }
    //判两直线垂直
    int perpendicular(line u,line v)
    {
        return zero((u.a.x-u.b.x)*(v.a.x-v.b.x)+(u.a.y-u.b.y)*(v.a.y-v.b.y));
    }
    //判两线段相交,包括端点和部分重合
    int intersect_in(line u,line v)
    {
        if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))
            return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);
        return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);
    }
 int find(int x)
 {
     if (f[x]!=x)
         f[x]=find(f[x]);
     return f[x];
 }

 void iunion(int x,int y)
 {
     int fx,fy;
     fx=find(x);
     fy=find(y);
     if (fx!=fy)
         f[fx]=fy;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     int times;
     cin>>times;
     while(times--)
     {

     cin>>n;
     int T=;
     for (int i=;i<=n;i++)
     {
         /*
         cout<<"dev: ";
         for (int j=1;j<=T;j++)
             cout<<f[j]<<" ";
         cout<<endl;
         */
         cin>>ch;
         if (ch=='Q')
         {
             cin>>tmp;
             int ans=;
             for (int j=;j<=T;j++)
                 if (find(tmp)==find(j))     ans++;
             cout<<ans<<endl;
         }
         else if (ch=='P')
         {
             cin>>X1>>Y1>>X2>>Y2;
             T++;
             f[T]=T;
             l[T]=line(point(X1,Y1),point(X2,Y2));
             //cout<<l[T].a.x<<" "<<l[T].a.y<<" "<<l[T].b.x<<" "<<l[T].b.y<<endl;
             for (int j=;j<T;j++)
                 if (intersect_in(l[T],l[j])>)
                     iunion(j,T);
         }
     }

     if (times>)
         cout<<endl;
     }
     return ;
 }