Segment set

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

分析：这道题是道基础带权并查集的题，主要就是如何判断两条线段相交（注意：是线段）

有一个链接：http://blog.csdn.net/rickliuxiao/article/details/6259322

为了确定两条线段是否相交，要检查每个线段是否跨越了包含另一线段的直线。 给定一个线段p1p2,如果点p1位于某一直线的一边， 而点p2位于直线的另一边，则称线段p1p2跨越了该直线。如果p1和p2 就落在该直线的话，即出现边界情况。两条线段相交， 当且仅当下面两个条件中有一个成立，或同时成立：

1)每个线段都跨越包含了另一线段的直线。

2)一个线段的某一端点位于另一线段上。

1. #include<stdio.h>
2. #include<string.h>
3. #include<algorithm>
4. #include<iostream>
5. #define Min(a,b) a<b?a:b
6. #define Max(a,b) a>b?a:b
7. #define maxh 1000+10
8. using namespace std;
9. typedef struct{
10.     double x,y;
11. }point;
12. typedef struct{
13.     point p0,p1;
14. }node;
15. node E[maxh];
16. int set[maxh],rank[maxh];
17. void init(){
18.     for(int i=0;i<maxh;i++){
19.         set[i]=i,rank[i]=1;
20.     }
21. }
22. int findx(int x){
23.     if(x==set[x])
24.     return x;
25.     set[x]=findx(set[x]);
26.     return set[x];
27. }
28. void Union(int x,int y){
29.     x=findx(x);
30.     y=findx(y);
31.     if(x!=y){
32.         set[x]=y;
33.         rank[y]+=rank[x];
34.     }
35. }
36. double direction(point p0,point p1,point p2){
37.     return ((p2.x-p0.x)*(p1.y-p0.y)-(p1.x-p0.x)*(p2.y-p0.y));
38. }
39. bool on(point p0,point p1,point p2){
40.     double xmin,ymin,xmax,ymax;
41.     xmin=Min(p0.x,p1.x);
42.     xmax=Max(p0.x,p1.x);
43.     ymin=Min(p0.y,p1.y);
44.     ymax=Max(p0.y,p1.y);
45.     if(p2.x>=xmin&&p2.x<=xmax&&p2.y>=ymin&&p2.y<=ymax)
46.     return true;
47.     return false;
48. }
49. bool intersect(point p0,point p1,point p2,point p3){
50.     double d0,d1,d2,d3;
51.     d0=direction(p2,p3,p0);
52.     d1=direction(p2,p3,p1);
53.     d2=direction(p0,p1,p2);
54.     d3=direction(p0,p1,p3);
55.   //  printf("%lf %lf %lf %lf\n",d0,d1,d2,d3);
56.     if(((d0<0&&d1>0)||(d0>0&&d1<0))&&((d2>0&&d3<0)||(d2<0&&d3>0))){
57.     //    printf("cwefg");
58.     return true;
59.     }
60.     else if(d0==0&&on(p2,p3,p0))
61.     return true;
62.     else if(d1==0&&on(p2,p3,p1))
63.     return true;
64.     else if(d2==0&&on(p0,p1,p2))
65.     return true;
66.     else if(d3==0&&on(p0,p1,p3))
67.     return true;
68.     else
69.     return false;
70. }
71. int main(){
72.     int T,n,h,x;
73.     char c[2];
74.     scanf("%d",&T);
75.     while(T--){
76.         scanf("%d",&n);
77.         init();
78.         h=1;
79.         while(n--){
80.             scanf("%s",c);
81.             if(c[0]==‘P‘){
82.                 scanf("%lf%lf%lf%lf",&E[h].p0.x,&E[h].p0.y,&E[h].p1.x,&E[h].p1.y);
83.                 for(int i=1;i<h;i++){
84.                     if(intersect(E[i].p0,E[i].p1,E[h].p0,E[h].p1)){
85.                    //     printf("<%d,%d>\n",i,h);
86.                         Union(i,h);
87.                     }
88.                 }
89.                 h++;
90.             }
91.             else{
92.                 scanf("%d",&x);
93.                 printf("%d\n",rank[findx(x)]);
94.             }
95.         }
96.         if(T){
97.             printf("\n");
98.         }
99.     }
100.     return 0;
101. }
102.                     
103.