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[1010];
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[1010];//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=0;
    for (int i=1;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=0;
            for (int j=1;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=1;j<T;j++)
                if (intersect_in(l[T],l[j])>0)
                    iunion(j,T);
        }
    }

    if (times>0)
        cout<<endl;
    }
    return 0;
}