poj 2524 Ubiquitous Religions

题目描述
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.

大致题意：有多组测试，每组测试首先输入两个数n，m，表示有n个学生编号1到n，接下来m行每行输入两个数x，y表示，编号为x和y的学生有相同的信仰。问最少有多少种信仰。
输入0 0结束

思路：简单的并查集

代码如下

#include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<map>using namespace std;int pre[50005];int total;int find(int x){    int r=x;   while (pre[r]!=r)   r=pre[r];   int i=x; int j;   while(i!=r)   {       j=pre[i];       pre[i]=r;       i=j;   }   return r;}void join(int x,int y){    int f1=find(x);    int f2=find(y);    if(f1!=f2)    {        pre[f2]=f1;        total--;    }}int main(){    int n,m;    int f=1;    while(1)    {        cin>>n>>m;        if(!(n+m)) break;        for(int i=0;i<=n;i++)        pre[i]=i;        total=n;        for(int i=1;i<=m;i++)        {            int x,y;            cin>>x>>y;            join(x,y);        }       printf("Case %d: %d\n",f,total);       f++;    }   return 0;}