hdu 2473 Junk-Mail Filter（删除节点操作的并查集）

Junk-Mail Filter

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5997    Accepted Submission(s): 1907

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

 

题意：

M：把两个数合并到同一个集合中，

S：把一个数从这个集合中删除（其他本来在这个集合中得数都仍属于这个集合，不管和这个数有没关联）

输出：集合的个数

思路（如何删除节点不影响该节点下面的点）：

把被删掉的那个点设为虚点，并新建一个点，把原来的点映射到这个新点上，用real[a]表示a映射过去的点，没有映射的仍是它本身。有了映射以后的操作都是对这个新点进行操作的。最后数一下根节点的个数就是集合数。

#include 
#include 
#include 
#include 
using namespace std;

int fa[1000010],real[1000010];//fa[ ]数组应该开到10^6
int vis[1000010],ans=0;

int find(int x)
{
	return x==fa[x]?x:fa[x]=find(fa[x]);
}

int main()
{
	int n,m,number=0;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==0 && m==0)
		{
			break;
		}
		for(int i=0;i