poj 2524 Ubiquitous Religions

最新推荐文章于 2019-09-06 17:22:50 发布

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博客给出POJ 2524题链接及描述，题目是通过询问学生对是否信仰相同宗教，求大学学生可能信仰的不同宗教最大数量。分析指出成对输入学生编号，相同编号代表信仰相同宗教，可合并为一个宗教，简单运用并查集求解。

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题目链接：http://poj.org/problem?id=2524

题目描述：

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

题目分析：成对的输入学生编号，每一对数字代表他们的宗教信仰是相同的，若在两组数据中有相同的数字编号则代表他们的宗教信仰是相同的，合并为1个宗教。这里就简单运用了并查集并未夹杂其他知识点。

代码：

#include<cstdio>
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=0x3f3f3f;
int a[50010];

int find(int x)
{
   return a[x]==x?x:(a[x]=find(a[x]));
}


int main()
{
    int n,m,I,J;
    int cnt=1;
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        int sum=n;//这里让宗教总数等于最大编号数，下方再根据条件自减
        for(int i=1;i<=n;i++)
            a[i]=i;
        for(int i=0;i<m;i++)
        {
          scanf("%d%d",&I,&J);
          I=find(a[I]);   //分别找各自对应的宗教
          J=find(a[J]);
          if(I!=J)      //每一对数首先保证这一对数不相同。由于不相同，故将他们的宗教合并，独立宗教少一个
          {
                a[I]=J;//路径压缩（即对应宗教相同的两个数，使两者对应的父节点相同）
                sum--;
          }

        }

       printf("Case %d: %d\n",cnt++,sum);
    }
    return 0;
}