# POJ 2524 Ubiquitous Religions C语言版

POJ题解 3490浏览

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

Slyar:说下题目大意。已知有n个大学生，其中有m对宗教信仰相同的学生，请你估算这n个学生中最多有多少种宗教信仰。

```#include
#include

/*
num[]存储节点所在集合元素的个数
father[]存储节点的父节点
*/
int father, num;

/* 初始化集合 */
void Make_Set(int x)
{
father[x] = x;
num[x] = 0;
}

/* 查找x元素所在的集合,回溯时压缩路径 */
int Find_Set(int x)
{
if (x != father[x])
{
father[x] = Find_Set(father[x]);
}
return father[x];
}

/* 合并a,b所在的集合 */
void Union_Set(int a, int b)
{
/* 如果两个元素在同一个集合则不需要合并 */
if (a == b) return;
/* 将小集合合并到大集合中，更新集合个数 */
if (num[a] <= num[b])
{
father[a] = b;
num[b] += num[a];
}
else
{
father[b] = a;
num[a] += num[b];
}
}

int main()
{
int i, j, n, m, a, b;
int t = 1;
while (1)
{
scanf("%d%d", &n, &m);
if (n + m == 0) break;
/* 将每个节点作为一个集合 */
for (i = 0; i < n; i++)
{
Make_Set(i);
}
for (i = 0; i < m; i++)
{
scanf("%d %d", &a, &b);
a = Find_Set(a);
b = Find_Set(b);
/* 不在同一集合则合并，n减1 */
if (a != b)
{
n--;
Union_Set(a, b);
}
}
printf("Case %d: %d\n", t++, n);
}
system("pause");
return 0;
}
```

### 网友最新评论 (2)

1. @苏洋, 不好...我没弄评论标签，不好玩...
Slyar12年前 (2009-05-27)回复
2. MakeSET直接放Sub Main 不好么...你的评论表情呢..消失还是没有存在过...
苏洋12年前 (2009-05-26)回复