并查集，在一些有N个元素的集合应用问题中，我们通常是在开始时让每个元素构成一个单元素的集合，然后按一定顺序将属于同一组的元素所在的集合合并，其间要反复查找一个元素在哪个集合中。
通过以链式的方式，实现顺序化。
例题
There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.
You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.
Input
The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.
Output
For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.
Sample Input
10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample Output
Case 1: 1
Case 2: 7
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
const int MX = 55555;
int stu[MX];
int n, m;
void ini() {
for (int i = 1; i <= n; i++) {
stu[i] = i;
}
}
int FindRoot(int pos) {
return stu[pos] == pos ? pos : (stu[pos] = FindRoot(stu[pos]));
}
int main() {
int sign = 1;
while (scanf("%d %d", &n, &m), n || m) {
ini();
int a, b;
for (int i = 0; i < m; i++) {
scanf("%d %d", &a, &b);
int root1 = FindRoot(a);
int root2 = FindRoot(b);
if (root1 != root2) {
n--;
stu[root2] = root1;
}
}
printf("Case %d: %d\n", sign++, n);
}
return 0;
}
为了节约时间，可以通过路径压缩的方法
int find(int x){
int son,temp;
son=x;
while(x!=pre[x]) x=pre[x];//找到最顶头的父亲
while(son!=x){
temp=pra[son];//现在son的父亲
pra[son]=x;//把现在的son的直接父亲变成最终父亲
son=temp;//准备把直接父亲的父亲变化
}
return x;
}
二.并查集
原文：https://www.cnblogs.com/BlogBaudelaire/p/14346291.html
