题目描述：

Junk-Mail Filter

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 620    Accepted Submission(s): 174

Problem Description
Recognizing junk mails is a tough task. The method used here consists of two steps:
1) Extract the common characteristics from the incoming email.
2) Use a filter matching the set of common characteristics extracted to determine whether the email is a spam.

We want to extract the set of common characteristics from the N sample junk emails available at the moment, and thus having a handy data-analyzing tool would be helpful. The tool should support the following kinds of operations:

a) “M X Y”, meaning that we think that the characteristics of spam X and Y are the same. Note that the relationship defined here is transitive, so
relationships (other than the one between X and Y) need to be created if they are not present at the moment.

b) “S X”, meaning that we think spam X had been misidentified. Your tool should remove all relationships that spam X has when this command is received; after that, spam X will become an isolated node in the relationship graph.

Initially no relationships exist between any pair of the junk emails, so the number of distinct characteristics at that time is N.
Please help us keep track of any necessary information to solve our problem.
 

Input
There are multiple test cases in the input file.
Each test case starts with two integers, N and M (1 ≤ N ≤ 10 5 , 1 ≤ M ≤ 10 6), the number of email samples and the number of operations. M lines follow, each line is one of the two formats described above.
Two successive test cases are separated by a blank line. A case with N = 0 and M = 0 indicates the end of the input file, and should not be processed by your program.
 

Output
For each test case, please print a single integer, the number of distinct common characteristics, to the console. Follow the format as indicated in the sample below.
 

Sample Input 
5 6
M 0 1
M 1 2
M 1 3
S 1
M 1 2
S 3

3 1
M 1 2

0 0
 

Sample Output 
Case #1: 3
Case #2: 2
 

Source
西湖之恋——HDU现场赛题目
 

Recommend
lcy

分析：
有于有可能要删掉以前的根节点。所以要保存它的信息。
a[i] 表是原来编号为i的字母的位置，例如原来是a[i]=i;
现在删掉i后，假如 i 原来是一个集合的根节点。那么，只对b[i]--;
即这个集合的个数 减一。而这个i  的位置则移动到数组最后面 a[sum]中。
此时a[i]=sum;完成i的转移。
因而当findroot(i)时应该变成 find(a[i])，
相当于寻址。

MY ACCEPTED CODE：
/**/ /*
******************************************************
HDU 2473 Junk-Mail Filter (2008Regional杭州现场赛题目)
using 并差集
memory 5384K
runtime 1265MS
codelength 1373B
Accepted time 2009-05-01 15:28:34
1327276 2009-05-01 15:28:34 Accepted 2473 1265MS 5384K 1373 B G++ baozi 
******************************************************
*/
#include < iostream >
using   namespace  std;
 #define  Maxn 1100005
int  father[Maxn],a[Maxn],b[Maxn];
 int  n,q;
 void  init()    
 {
    for(int i=0;i<=n;i++)
    {
        father[i]=i;
        a[i]=i;
        b[i]=1;    
    }    
}
int  findroot( int  x)
 {
    if(x!=father[x])
     father[x]=findroot(father[x]);
    return father[x];
}
void  UNION( int  x, int  y)
 {
    b[y]+=b[x];
    b[x]=0;
    father[x]=y;
}
int  main()
 {
    int i,ans,count=1,sum,x,y,fa,fb;
    char que[20];
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        if(n==0&&q==0) break;
        init(); 
        sum=n;   
        for(i=0;i<q;i++)
        {
            scanf("%s",que);
            if(strcmp(que,"M")==0)
            {
                scanf("%d%d",&x,&y);
                fa=findroot(a[x]); 
                fb=findroot(a[y]); 
                if(fa!=fb)
                UNION(fa,fb);  
            }
            else
            {
                scanf("%d",&x);
                fa=findroot(a[x]);
                b[fa]--;
                b[sum]=1;
                a[x]=sum;
                father[sum]=sum;
                sum++;
            }
           
        }
            ans=0;
            for(i=0;i<sum;i++)
            if(b[i]>0)
             ans++;
            printf("Case #%d: %d\n",count++,ans);
    }
    
    return 0;    
}