hdu 1558 线段相交+并查集路径压缩

Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3457    Accepted Submission(s): 1290

Problem Description

A segment and all segments which are connected with it compose a segment set. The size of a segment set is the number of segments in it. The problem is to find the size of some segment set.

 

Input

In the first line there is an integer t - the number of test case. For each test case in first line there is an integer n (n<=1000) - the number of commands.

There are two different commands described in different format shown below:

P x1 y1 x2 y2 - paint a segment whose coordinates of the two endpoints are (x1,y1),(x2,y2).
Q k - query the size of the segment set which contains the k-th segment.

k is between 1 and the number of segments in the moment. There is no segment in the plane at first, so the first command is always a P-command.

 

Output

For each Q-command, output the answer. There is a blank line between test cases.

 

Sample Input

1

10

P 1.00 1.00 4.00 2.00

P 1.00 -2.00 8.00 4.00

Q 1

P 2.00 3.00 3.00 1.00

Q 1

Q 3

P 1.00 4.00 8.00 2.00

Q 2

P 3.00 3.00 6.00 -2.00

Q 5

 

Sample Output

1
2
2
2
5

 

Author

LL

 

Source

HDU 2006-12 Programming Contest

 

题目大意：有n个指令，p加入一条线段,q查询id线段所在集合（两线段有交点为同一集合）的元素个数。

思路：用并查集路径压缩记录各个线段间的关系，根据叉积的定义有：Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方。

两线段有交点的必要条件：必须每条线段的两个端点在另一线段的两侧或直线上。

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

 const double eps=1e-;
 const int maxn=;
 int f[maxn];
 struct Point
 {
     double x,y;
     Point(){}
     Point(double x,double y):x(x),y(y){}
 };
 struct Line
 {
     Point a,b;
 }L[maxn];
 typedef Point Vector;
 Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}//叉积

 bool judge(Line a,Line b)//Cross(v,w)=0时w在v上,>0时w在v上方,<0时w在v下方
 {
     if(dcmp(Cross(a.a-b.a,b.b-b.a)*Cross(a.b-b.a,b.b-b.a))<=
         &&dcmp(Cross(b.a-a.a,a.b-a.a)*Cross(b.b-a.a,a.b-a.a))<=)
         return true;
     return false;
 }
 int findset(int x){return f[x]!=x?f[x]=findset(f[x]):x;}
 void Union(int a,int b)
 {
     a=findset(a);b=findset(b);
     if(a!=b) f[a]=b;
 }
 int main()
 {
     int t,n,i,j,id;
     char op[];
     double x1,y1,x2,y2;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         int cnt=;
         for(i=;i<n;i++)
         {
             scanf("%s",op);
             if(op[]=='P')
             {
                 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
                 L[++cnt].a=Point(x1,y1);L[cnt].b=Point(x2,y2);
                 f[cnt]=cnt;
                 for(j=;j<=cnt-;j++)
                     if(judge(L[cnt],L[j]))
                         Union(j,cnt);
             }
             else
             {
                 int ans=;scanf("%d",&id);
                 id=findset(id);
                 for(j=;j<=cnt;j++)
                     if(findset(j)==id)
                         ans++;
                 printf("%d\n",ans);
             }
         }
         if(t) printf("\n");
     }
     return ;
 }