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 }
ACM1558两线段相交判断和并查集,布布扣,bubuko.com