这道题是并查集简单题,题目如下：
Description
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
Hint
Huge input, scanf is recommended.
Source
Alberta Collegiate Programming Contest 2003.10.18
题目思路：将信仰同一个宗教的几个人归于同一个集合，最终集合个数即为答案。AC代码如下：
与这道题类似的题目还有hdu 1213 "How Many Tables"
#include<iostream> #include<stdio.h> using namespace std; const int maxn=50001; int s[maxn],height[maxn]; void init_set(){//初始化 for(int i=1;i<=maxn;i++) s[i]=i,height[i]=0; } int find_set(int x){ return x==s[x]?x:find_set(s[x]); }//查找根节点 void union_set(int x,int y){//优化合并操作 x=find_set(x); y=find_set(y); if(height[x]==height[y]){ height[x]++;//两个树的高度相同，将其合并，数的高度加一 s[y]=x; } else { if(height[x]<height[y]) s[x]=y;//将矮树并到高树上，高树的高度保持不变 else s[y]=x; } } int main(){ int n,m,x,y; int c=1; while(~scanf("%d %d",&n,&m)){ if(n==0&&m==0) break; init_set(); for(int i=1;i<=m;i++){ scanf("%d %d",&x,&y); union_set(x,y); } int count=0; for(int i=1;i<=n;i++) if(s[i]==i) count++; printf("Case %d: %d\n",c++,count); }return 0; }