Arbitrary point ~ Union search to find connected blocks

Link: https://www.nowcoder.com/acm/contest/84/C
Source: Niuke.com

any point

Time limit: C/C++ 1 second, other languages ​​2 seconds
Space limit: C/C++ 32768K, other languages ​​65536K
64bit IO Format: %lld

Topic description

There are several points on the plane. From each point, you can walk in any direction, until you hit another point, and then you can change the direction.
How many points need to be added at least so that the point pairs can reach each other.

Enter description:

The first line contains an integer n representing the number of points ( 1 <= n <= 100). 
The second line is n lines, each with two integers x

_i

, and

_i

represents the coordinates ( 1 <= x

_i

, and

_i 

<= 1000). 
The positive direction of the y-axis is north, and the square of the x-axis is east.

Output description:

Output an integer indicating the minimum number of points to add.

Example 1

enter

2
2 1
1 2

output

1

Example 2

enter

2
2 1
4 1

output

0 

This is a parallel search set for Unicom Express

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <vector>
 4 #include <queue>
 5 #include <cstring>
 6 #include <string>
 7 using namespace std;
 8 const int maxn = 1e4 + 10;
 9 
10 typedef long long LL ;
11 int fa[maxn], x[maxn], y[maxn];
12 
13 void init(int n) {
14     for (int i = 0 ; i <= n ; i++) {
15         fa[i] = i;
16     }
17 }
18 int find(int p) {
19     while(p != fa[p]) p = fa[p];
20     return p;
21 }
22 int find_set(int p)
23 {
24     while(p!=fa[p]) p=fa[p];
25     return p;
26 }
27 
28 void combine(int p, int q) {
29     int np = find_set(p);
30     int nq = find_set(q);
31     if (np != nq) fa[np] = nq;
32 }
33 
34 int main() {
35     int n;
36     //freopen("DATA.txt","r",stdin);
37     while(scanf("%d", &n) != EOF) {
38         init(n);
39         for (int i = 1 ; i <= n ; i++) {
40             scanf("%d%d", &x[i], &y[i]);
41         }
42         for (int i = 1 ; i < n ; i++)
43             for (int j = i + 1 ; j <= n ; j++)
44                 if (x[i] == x[j] || y[i] == y[j] ) combine(i, j);
45         int ans = 0;
46         for (int i = 1 ; i <= n ; i++)
47             if (fa[i] == i) ans++;
48         printf("%d\n", years - 1 );
49      }
 50      return  0 ;
51 }