POJ 2524 并查集

Ubiquitous Religions
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 20197   Accepted: 9920

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这是一道水题，基本就直接套模板就行了，不需要被别的什么东西，尼玛，结果就这样一个破题，因为模板敲错了，结果就wa了半天，，最后被小bb一眼看出了错误，接着才过，下面是代码：
 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn=10000000;
int pre[maxn];

int find(int v){
   return pre[v]==v?v:find(pre[v]);
}

void join(int a,int b){
     int f1=find(a),f2=find(b);
     if(f1!=f2)
       pre[f1]=f2;
}

int main(){
    int n,m;
    int count;
for(count=1;;count++){
    cin>>n>>m;
    if(n==0&&m==0)
       break;
    for(int i=1;i<=n;i++){
          pre[i]=i;
    }
    int max=n;
    for(int i=1;i<=m;i++){
          int a,b;
          cin>>a>>b;
          if(find(a)!=find(b))
              max=max-1;
          join(a,b);
     }
      printf("Case %d: ",count);
      printf("%d\n",max);
}
    return 0;
}

补充：软件开发 , C++ ,

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