poj 2524

通过率：47% 难度：一般
Ubiquitous Religions
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 11902 Accepted: 5609 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
翻译：现如今世界上有如此多的宗教信仰，要将它们所有的都保持联系是非常困难的。

你对于找出在一个大学里面同学们有多少不同的宗教信仰感兴趣。

你知道这个大学里面有n个学生。

如果让你去一个一个问每一个学生的宗教信仰是不可能的。

而且，许多学生不愿意表露他们的宗教信仰。

有个方法可以避免这个问题，就是去问m （0<=m<=n(n-1)/2）对学生，并且问它们是否同一个宗教信仰（比如说如果他们是同一个宗教信仰，他们也许相互认识）。

对于这个数据，你可能并不知道吗，每一个学生的信仰，但是你可以知道最多有多少个不同的宗教信仰在这个校园里面。

你可以假设每一个学生有至多一个宗教信仰
输入数据包括情况的数目，每一种情况包括有数据n和m，接下来的m行包括了两个数据i和j，表示学生i和j是同一个宗教信仰。

学生是从1到n编号的，若n=0并且m=0则表示输入结束
对于每一种情况，显示出在这个学校里学生们的最大的不同宗教信仰数
解题思路：该题主要就是用到一个并查集，对于每一种情况，假设最初有num=-n（n表示访问学生数）种宗教信仰，然后使用并查集中，没合并一次就在num上减去1，最后输出num即可
总结及出错情况：该题较为简单，关键用好并查集就好，不易出错
代码：
#include<stdio.h>
#include<stdlib.h>
int father[50005];
void Make_set(int x)
{
father[x]=x;
}
int Find_set(int x)
{
if(father[x]==x)
return x;
father[x]=Find_set(father[x]);
return father[x];
}
void Union_set(int x,int y)
{
int rx=Find_set(x);
int ry=Find_set(y);
if(rx!=ry)
{
father[ry]=rx;
}
}
int main()
{
int n,m,i,j,num,sum=0,x,y,rx,ry;
while(scanf("%d%d",&n,&m))
{
if(n==0&&m==0)
break;
num=n;
sum=sum+1;
for(i=1;i<=n;i++)
{
Make_set(i);
}
for(i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
rx=Find_set(x);
ry=Find_set(y);
if(rx!=ry)
{
num=num-1;
Union_set(x,y);
}
}
printf("Case %d: %d\n",sum,num);
}
system("pause");
return 0;
}。