HDU 1558 Segment set （并查集+线段非规范相交）

题目链接

题意 ： 如果两个线段相交就属于同一集合，查询某条线段所属集合有多少线段，输出。

思路 ： 先判断与其他线段是否相交，然后合并。

 //
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <cmath>
 #define eps 1e-8
 #define zero(x) (((x) > 0 ? (x) : (-x)) < eps)

 using namespace std ;

 struct point
 {
     double x,y ;
 } p[];
 struct Line
 {
     point a;
     point b ;
 //    int num ;
 } L[] ;
 int father[],numb[],rankk[] ;
 int  cnt ;
 double direction(point p0,point p1,point p2)
 {
     return (p2.x-p0.x)*(p1.y-p0.y)-(p1.x-p0.x)*(p2.y-p0.y);
 }

 bool on_segment(point p0,point p1,point p2)
 {
     if((min(p0.x,p1.x)<=p2.x && p2.x<=max(p0.x,p1.x)) && (min(p0.y,p1.y)<=p2.y && p2.y<=max(p0.y,p1.y)))
         return true;
     return false;
 }

 bool Segment_intersect(point p1,point p2,point p3,point p4)
 {
     double d1,d2,d3,d4;
     d1 = direction(p3,p4,p1);
     d2 = direction(p3,p4,p2);
     d3 = direction(p1,p2,p3);
     d4 = direction(p1,p2,p4);
     if(((d1> && d2<)||(d1< && d2>)) && ((d3> && d4<)||(d3<&&d4>)))
         return true;
     else if((d1== && on_segment(p3,p4,p1)) || (d2== && on_segment(p3,p4,p2)) || (d3== && on_segment(p1,p2,p3)) || (d4== && on_segment(p1,p2,p4)))
         return true;
     return false;
 }
 int find_(int x)
 {
     if(father[x] != x)
         father[x] = find_(father[x]) ;
     return father[x] ;
 }

 void mergee(int a, int b){
      int fx = find_(a);
      int fy = find_(b);

      if (fx != fy){
            father[fy] = fx;
            numb[fx] += numb[fy];
      }
 }
 void Init()
 {
     cnt =  ;
     for(int i=; i<=; i++)
     {
         numb[i]=;
     }
     for(int i =  ; i <  ; i++)
         father[i] = i ;
     memset(rankk,,sizeof(rankk)) ;
 }
 int main()
 {
     int T ,n,ss;
     scanf("%d",&T) ;
     char s[] ;
     while( T--)
     {
         scanf("%d",&n) ;
         Init() ;
         for(int i =  ; i < n ; i++)
         {
             scanf("%s",s) ;
             if(s[] == 'P')
             {
                 cnt ++ ;
                 scanf("%lf %lf %lf %lf",&L[cnt].a.x,&L[cnt].a.y,&L[cnt].b.x,&L[cnt].b.y) ;
                 for(int j =  ; j < cnt ; j ++)
                     if(find_(j) != find_(cnt) && Segment_intersect(L[j].a,L[j].b,L[cnt].a,L[cnt].b))
                         mergee(j,cnt) ;
             }
             else
             {
                 scanf("%d",&ss) ;
                 printf("%d\n",numb[find_(ss)]) ;
             }
         }
         if(T) printf("\n") ;
     }
     return  ;
 }

线段非规范相交1

 double cross(point p0,point p1,point p2)
 {
     return (p2.x-p0.x)*(p1.y-p0.y)-(p1.x-p0.x)*(p2.y-p0.y);
 }

 bool on_segment(point p0,point p1,point p2)
 {
     if((min(p0.x,p1.x)<=p2.x && p2.x<=max(p0.x,p1.x)) && (min(p0.y,p1.y)<=p2.y && p2.y<=max(p0.y,p1.y)))
         return true;
     return false;
 }

 bool Segment_intersect(point p1,point p2,point p3,point p4)
 {
     double d1,d2,d3,d4;
     d1 = cross(p3,p4,p1);
     d2 = cross(p3,p4,p2);
     d3 = cross(p1,p2,p3);
     d4 = cross(p1,p2,p4);
     if(((d1> && d2<)||(d1< && d2>)) && ((d3> && d4<)||(d3<&&d4>)))
         return true;
     else if((d1== && on_segment(p3,p4,p1)) || (d2== && on_segment(p3,p4,p2)) || (d3== && on_segment(p1,p2,p3)) || (d4== && on_segment(p1,p2,p4)))
         return true;
     return false;
 }

线段非规范相交2

 double cross(point a, point b, point c)
 {
     return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
 }

 //aa, bb为一条线段两端点 cc, dd为另一条线段的两端点 相交返回true, 不相交返回false
 bool intersect(point aa, point bb, point cc, point dd)
 {
     if ( max(aa.x, bb.x)<min(cc.x, dd.x) )
     {
         return false;
     }
     if ( max(aa.y, bb.y)<min(cc.y, dd.y) )
     {
         return false;
     }
     if ( max(cc.x, dd.x)<min(aa.x, bb.x) )
     {
         return false;
     }
     if ( max(cc.y, dd.y)<min(aa.y, bb.y) )
     {
         return false;
     }
     if ( cross(cc, bb, aa)*cross(bb, dd, aa)< )
     {
         return false;
     }
     if ( cross(aa, dd, cc)*cross(dd, bb, cc)< )
     {
         return false;
     }
     return true;
 }