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：

  1 #include<iostream>
  2 using namespace std;
  3 #define MIN(x,y) (x < y ? x : y)
  4 #define MAX(x,y) (x > y ? x : y)
  5 typedef struct 
  6 {
  7     double x,y;
  8 } Point;
  9 struct LINE
 10 {
 11     double x1,x2,y1,y2;
 12     int flag;    
 13 };
 14 LINE *line;
 15 int lineintersect(int a,int b)
 16 {
 17     Point p1,p2,p3,p4;
 18     p1.x=line[a].x1;p1.y=line[a].y1;
 19     p2.x=line[a].x2;p2.y=line[a].y2;
 20     p3.x=line[b].x1;p3.y=line[b].y1;
 21     p4.x=line[b].x2;p4.y=line[b].y2;
 22     
 23     Point tp1,tp2,tp3,tp4,tp5,tp6;
 24     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))
 25         return 1;
 26 //快速排斥试验
 27 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))
 28             return 0;
 29 //跨立试验
 30     tp1.x=p1.x-p3.x;
 31     tp1.y=p1.y-p3.y;
 32     tp2.x=p4.x-p3.x;
 33     tp2.y=p4.y-p3.y;
 34     tp3.x=p2.x-p3.x;
 35     tp3.y=p2.y-p3.y;//从这到上面是一组的向量
 36     tp4.x=p2.x-p4.x;//下面是一组的向量
 37     tp4.y=p2.y-p4.y;
 38     tp5.x=p2.x-p3.x;
 39     tp5.y=p2.y-p3.y;
 40     tp6.x=p2.x-p1.x;
 41     tp6.y=p2.y-p1.y;
 42     if ((tp1.x*tp2.y-tp1.y*tp2.x)*(tp2.x*tp3.y-tp2.y*tp3.x)>=0) //此处用到了叉乘公式
 43     {
 44         if((tp4.x*tp6.y-tp4.y*tp6.x)*(tp6.x*tp5.y-tp6.y*tp5.x)>=0)
 45             return 1;  
 46     }
 47 return 0;
 48 }
 49 int find(int x)
 50 {
 51     if(0>line[x].flag)return x;
 52     return line[x].flag=find(line[x].flag);
 53 }
 54 void Union(int a,int b)
 55 {
 56     int fa=find(a);
 57     int fb=find(b);
 58     if(fa==fb)return;
 59     int n1=line[fa].flag;
 60     int n2=line[fb].flag;
 61     if(n1>n2)
 62     {
 63         line[fa].flag=fb;
 64         line[fb].flag+=n1;
 65     }
 66     else 
 67     {
 68         line[fb].flag=fa;
 69         line[fa].flag+=n2;
 70     }
 71 }
 72 void throwinto(int n)
 73 {
 74     if(n>1)
 75     {
 76         for(int i=1;i<n;i++)
 77         {
 78             if(lineintersect(i,n))
 79             Union(i,n);
 80         }
 81     }
 82 }
 83 int main()
 84 {
 85     int t;
 86     cin>>t;
 87     int n,q;
 88     int cn=t;
 89     while(t--)
 90     {
 91         
 92         int cnt=1;
 93         cin>>n;
 94         line=new LINE[n+1];
 95         for(int i=0;i<=n;i++)
 96         {
 97             line[i].flag=-1;
 98         }
 99         for(int i=1;i<=n;i++)
100         {
101             char sj;
102             cin>>sj;
103             if(sj=='P')
104             {
105                 cin>>line[cnt].x1>>line[cnt].y1>>line[cnt].x2>>line[cnt].y2;
106                 throwinto(cnt);
107                 cnt++;
108             }
109             else if(sj=='Q')
110             {
111                 cin>>q;
112                 int re=find(q);
113                 cout<<-line[re].flag<<endl;
114             }
115         }
116         if(t>0)cout<<endl;//注意格式问题，容易出错
117     }
118     return 0;
119 }

 

posted @ 2014-07-29 22:32  SYTM  阅读(434)  评论(0)  编辑  收藏  举报