POJ 2524 Ubiquitous Religions(并查集)

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 16083		Accepted: 7764

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

Hint

Huge input, scanf is recommended.

Source

Alberta Collegiate Programming Contest 2003.10.18

 

解题报告：题意就是每个人是不同地区的，给出人数，和他们之间的关系，统计这些人所在不同区域的个数，其实就是统计一下集合的个数，利用并查集的算法就能实现这一功能；

代码如下：

#include <iostream>
#include <cstdio>
using namespace std;
const int N = 50010;
int father[N], ans;
void Makeset() //初始化函数
{
    int i;
    for (i = 0; i < N; ++i)
    {
        father[i] = i;
    }
}
int Find(int x) //找集合的“根”
{
    if (father[x] != x)
    {
        father[x] = Find(father[x]);
    }
    return father[x];
}
void Merge(int x, int y) //若不在一个集合中则合并到一个集合，若在一个集合中则不处理
{
    int i, j;
    i = Find(x);
    j = Find(y);
    if (i != j)//不在一个集合中的处理
    {
        ans --; //集合的数目减1
        if (i > j)
        {
            father[i] = j;
        }
        else
        {
            father[j] = i;
        }
    }
}
int main()
{
    int n, m, i, p, q, count;
    count = 0;
    while (scanf("%d%d", &n, &m) != EOF && n && m)
    {
        count ++;
        ans = n;//假设集合的个数最多
        Makeset();
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d", &p, &q);
            Merge(p, q);
        }
        printf("Case %d: %d\n", count, ans);//ans统计的是集合的个数
    }
    return 0;
}