POJ 2524 Ubiquitous Religions（中文对应）

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Ubiquitous Religions（宗教信仰）

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 30250		Accepted: 14639

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

Sample Output

Case 1: 1
Case 2: 7

Hint

Huge input, scanf is recommended.

Source

Alberta Collegiate Programming Contest 2003.10.18

中文版：

当今世界有很多不同的宗教,很难通晓他们。你有兴趣找出在你的大学里有多少种不同的宗教信仰。

你知道在你的大学里有n个学生(0 < n <= 50000) 。你无法询问每个学生的宗教信仰。此外,许多学生不想说出他们的信仰。避免这些问题的一个方法是问m(0 <= m <= n(n - 1)/ 2)对学生, 问他们是否信仰相同的宗教( 例如他们可能知道他们两个是否去了相同的教堂) 。在这个数据中,你可能不知道每个人信仰的宗教，但你可以知道校园里最多可能有多少个不同的宗教。假定每个学生最多信仰一个宗教。

Input

有多组数据。对于每组数据：

第一行：两个整数n和m。

以下m行：每行包含两个整数i和j，表示学生i和j信仰相同的宗教。学生编号从1到n。

输入的最后一行中,n = m = 0。

Output

对于每组测试数据，输出一行，输出数据序号( 从1开始) 和大学里不同宗教的最大数量。（参见样例）

简单的并查集问题，就是预习预习吧

代码：

#include<stdio.h>
#define MYDD 1103int pre[MYDD*50];//记录父节点
void init(int x) { //初始化for(int j=1; j<=x; j++) {pre[j]=j;//父节点等于本身}
}int find(int x) { //查找父节点return pre[x]==x? x:find(pre[x]);
}void combine(int x,int y) {//合并节点int fx=find(x);int fy=find(y);if(fx!=fy)pre[fx]=fy;
}int main() {int n,m,v=1;while(scanf("%d%d",&n,&m)&&(n||m)) {init(n);//初始化int a,b;for(int j=1; j<=m; j++) {scanf("%d%d",&a,&b);combine(a,b);}int ans=0;for(int j=1; j<=n; j++) {if(pre[j]==j)ans++;//父节点等于本身}printf("Case %d: %d\n",v++,ans);}return 0;
}

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