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 ≤ 105 , 1 ≤ M ≤ 106), 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
大致题意：
有n封邮件现在要将其含有相同的特征的放在一起，还要删掉一些误判的；
现在问你这两种操作完成后还有多少种的信。
解题思路：
主要是并查集的删点操作，建立虚拟节点。
1 #include <cstdio>
2 #include <cstring>
3 using namespace std;
4 int f[2*1000005],flag[4*1000005],x,y,ans,cnt,n,m,k=1; char c;
5 int sf(int x){ return x==f[x]? x:f[x]=sf(f[x]); }
6 int main(){
7 while(scanf("%d%d",&n,&m),n+m){
8 ans=0,cnt=2*n;
9 for(int i=0;i<n;i++) f[i]=n+i;
10 for(int i=n;i<2*n+m;i++) f[i]=i;
11 memset(flag,0,sizeof(flag));
12 for(int i=1;i<=m;i++){
13 scanf(" %c",&c);
14 if(c==‘M‘){
15 scanf("%d%d",&x,&y);
16 f[sf(x)]=sf(y);
17 } else {
18 scanf("%d",&x);
19 f[x]=cnt++;
20 }
21 }
22 for(int i=0;i<n;i++)
23 if(flag[sf(i)]==0){
24 ans++;
25 flag[sf(i)]=1;
26 } printf("Case #%d: %d\n",k++,ans);
27 } return 0;
28 }