poj 2524 Ubiquitous Religions 一简单并查集

Ubiquitous Religions

 

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 22389		Accepted: 11031

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.

讲解：并查集很好玩啊，哈哈。。。

        题目的大体意思是说，这个学校有n个人，编号1 到 n，下面输入 m 行，每行两个人，两个人在一个种族中；如果有人不在这个学校中，即编号大于n，则忽略不计；

        学校想给每个种族发一种东西，所以你要统计共有多少个种族，如果此人没有出现，则可认为他自己是一个种族，因为要求最大的；

        总共的就是 种族数加上没出现的人的个数；

        算是一个入门级的并查集应用了，比食物链那道题要简单多了，

上面是 scanf， 下面是  cin  时间上的差别很明显啊

AC代码：

 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<set>
 #include<cstdio>
 using namespace std;
 int father[];
 int vis[];//用来标记出现过没有；
 void begin(int m)
 {
     for(int i=;i<=m;i++)
         {father[i]=i;vis[i]=;}

 }
 int find(int x)
 {
     if(father[x]!=x)
     {
         father[x]=find(father[x]);
     }
     return father[x];
 }
 int main()
 {
     int m,n,x,y,a,b,mm,ans,k=;
     while(scanf("%d %d",&m,&n) && m+n)
     {   ans=;mm=;       //mm 统计的是出现的个数，减去就是没出现的个数；
         begin(m);
         for(int i=;i<=n;i++)
         {
             scanf("%d %d",&x,&y);//输入尽量用scanf（344ms），cin（4688ms）差别略大啊；
             if(x<=m && y<=m)
             {
                 a=find(x);b=find(y);//各找各的父亲；
                    father[a]=b;
                     if(vis[x]==)   //标记统计出现过多少个数；
                         {mm++;vis[x]=;}
                         if(vis[y]==)
                            {mm++;vis[y]=;}
             }
         }
         for(int i=;i<=m;i++)              //统计共有几个集合，实在不好想，用了个笨方法；
         {
               if(father[i]==i && vis[i]==)//父亲的父亲，是他本身，并且他已经出现过了
                   ans++;
         }
         printf("Case %d: %d\n",k++,ans+m-mm);
     }
     return ;
 }