Ubiquitous Religions(並查集的簡單應用)

題目來源：[NWPU][2014][TRN][12]並查集 E 題

http://vjudge.net/contest/view.action?cid=50731#problem/E

作者：npufz

題目：

E - Ubiquitous Religions

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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.

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int parent[50005];
int sum[50005];
int getparent(int a)
{
    if(a==parent[a])
        return a;
    int t;
    t=getparent(parent[a]);
    parent[a]=t;
    return  parent[a];
}
int mergege(int a,int b)
{
    int pre1,pre2;
    pre1=getparent(a);
    pre2=getparent(b);
    if(pre1==pre2)  return 0;
    if(sum[pre1]>sum[pre2])
    {
        parent[pre2]=pre1;
        sum[pre1]+=sum[pre2];
        return 0;
    }
    parent[pre1]=pre2;
    sum[pre2]+=sum[pre1];
    return 0;
}
int main()
{
    int n,cas,i,j,k,cnt=1,stu1,stu2,ma,pre,prea;
    while(scanf("%d%d",&n,&cas),n!=0&&cas!=0)
    {
        for(i=1;i<=n;i++)
        {
            parent[i]=i;
            sum[i]=1;
        }
        for(i=0;i<cas;i++)
        {
            scanf("%d%d",&stu1,&stu2);
            mergege(stu1,stu2);
        }
        ma=n;
        bool flag=true;
        //cout<<"ok"<<endl;
        for(i=1;i<=n;i++)
        {   //cout<<"ok"<<endl;
            pre=getparent(i);
            if(flag)
            {
                ma=ma-sum[pre]+1;
                prea=pre;
                flag=false;
            }
            else
            {
                if(pre!=prea)
                {
                   ma=ma-sum[pre]+1;
                   parent[pre]=prea;
                }
            }
       }
      printf("Case %d: %d\n",cnt,ma);
      cnt++;
 }
return 0;
}

反思：並查集可以把屬於同一組的元素合併到一起，並可以通過兩個集合的元素的關係，來把兩個集合合併到一起