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.


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.


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


1、将 a a a, b b b 所在的集合合并
2、将 a a a 移出它所在的集合


怎么将 a a a 移出集合呢?
如果 a a a 不是 父节点, 每次显然只需 f [ a ] = a f[a] = a f[a]=a一句话就行了。
如果 a a a 是父节点,我们很难维持集合中其他点的关系。
所以我们的想法就是让 a a a 一直不是父节点。我们只要每次建一个虚的点就可以了!这样操作起来简单方便快捷!


#include <bits/stdc++.h>
#define LL long long
#define N 2000005
using namespace std;
int f[N], t, cnt, s, v[N];
char o[3];
int find(int x){
	return x == f[x]? x: f[x] = find(f[x]);
int main(){
	int i, j, n, m, a, b;
	while(scanf("%d%d", &n, &m) && n){
		memset(v, 0, sizeof(v));
		cnt = 0;
		s = n;
		for(i = 1; i <= m + n; i++) f[i] = i;
			scanf("%s", o);
			if(o[0] == 'M'){
				scanf("%d%d", &a, &b);
				a++, b++;
				a = find(a); b = find(b);
				if(a <= n && b <= n){
					f[a] = f[b] = ++s;
				else if(a > n) f[b] = a;
				else if(b > n) f[a] = b;
				scanf("%d", &a);
				f[a] = a;
		for(i = 1; i <= n; i++){
				v[find(i)] = 1;
		printf("Case #%d: %d\n", t, cnt);
	return 0;