ACM1558两线段相交判断和并查集

Segment set

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

题意是要求输入一些线段，判断这些线段连接成几个集合，然后求包含某条线段的集合中的元素个数；

本题主要难关是判断量条线段是否相交，在我这篇文章里有详细介绍判断两条线段相交的方法。http://www.cnblogs.com/sytu/articles/3876585.html;

Notice:二维向量的叉乘公式是 axb=(x1,y1)x(x2,y2)=x1*y2-y1*x2;

下面为AC代码：

 #include<iostream>
 using namespace std;
 #define MIN(x,y) (x < y ? x : y)
 #define MAX(x,y) (x > y ? x : y)
 typedef struct
 {
     double x,y;
 } Point;
 struct LINE
 {
     double x1,x2,y1,y2;
     int flag;
 };
 LINE *line;
 int lineintersect(int a,int b)
 {
     Point p1,p2,p3,p4;
     p1.x=line[a].x1;p1.y=line[a].y1;
     p2.x=line[a].x2;p2.y=line[a].y2;
     p3.x=line[b].x1;p3.y=line[b].y1;
     p4.x=line[b].x2;p4.y=line[b].y2;

     Point tp1,tp2,tp3,tp4,tp5,tp6;
     if ((p1.x==p3.x&&p1.y==p3.y)||(p1.x==p4.x&&p1.y==p4.y)||(p2.x==p3.x&&p2.y==p3.y)||(p2.x==p4.x&&p2.y==p4.y))
         return ;
 //快速排斥试验
 if(MAX(p1.x,p2.x)<MIN(p3.x,p4.x)||MAX(p3.x,p4.x)<MIN(p1.x,p2.x)||MAX(p1.y,p2.y)<MIN(p3.y,p4.y)||MAX(p3.y,p4.y)<MIN(p1.y,p2.y))
             return ;
 //跨立试验
     tp1.x=p1.x-p3.x;
     tp1.y=p1.y-p3.y;
     tp2.x=p4.x-p3.x;
     tp2.y=p4.y-p3.y;
     tp3.x=p2.x-p3.x;
     tp3.y=p2.y-p3.y;//从这到上面是一组的向量
     tp4.x=p2.x-p4.x;//下面是一组的向量
     tp4.y=p2.y-p4.y;
     tp5.x=p2.x-p3.x;
     tp5.y=p2.y-p3.y;
     tp6.x=p2.x-p1.x;
     tp6.y=p2.y-p1.y;
     if ((tp1.x*tp2.y-tp1.y*tp2.x)*(tp2.x*tp3.y-tp2.y*tp3.x)>=) //此处用到了叉乘公式
     {
         if((tp4.x*tp6.y-tp4.y*tp6.x)*(tp6.x*tp5.y-tp6.y*tp5.x)>=)
             return ;
     }
 return ;
 }
 int find(int x)
 {
     if(>line[x].flag)return x;
     return line[x].flag=find(line[x].flag);
 }
 void Union(int a,int b)
 {
     int fa=find(a);
     int fb=find(b);
     if(fa==fb)return;
     int n1=line[fa].flag;
     int n2=line[fb].flag;
     if(n1>n2)
     {
         line[fa].flag=fb;
         line[fb].flag+=n1;
     }
     else
     {
         line[fb].flag=fa;
         line[fa].flag+=n2;
     }
 }
 void throwinto(int n)
 {
     if(n>)
     {
         for(int i=;i<n;i++)
         {
             if(lineintersect(i,n))
             Union(i,n);
         }
     }
 }
 int main()
 {
     int t;
     cin>>t;
     int n,q;
     int cn=t;
     while(t--)
     {

         int cnt=;
         cin>>n;
         line=new LINE[n+];
         for(int i=;i<=n;i++)
         {
             line[i].flag=-;
         }
         for(int i=;i<=n;i++)
         {
             char sj;
             cin>>sj;
             if(sj=='P')
             {
                 cin>>line[cnt].x1>>line[cnt].y1>>line[cnt].x2>>line[cnt].y2;
                 throwinto(cnt);
                 cnt++;
             }
             else if(sj=='Q')
             {
                 cin>>q;
                 int re=find(q);
                 cout<<-line[re].flag<<endl;
             }
         }
         if(t>)cout<<endl;//注意格式问题，容易出错
     }
     return ;
 }