clu info std -s ase ive trac can tinc

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

題意：S是將該點從集合中剝離出來，M是聯合起來。求出最後的聯通塊
思路：就是並查集的刪除操作了！多開一個數組來記錄當前點的映射位置！由於眼下還沒有其它的算法能將並查集的節點分離出來而不影響其結構。所以分離出來的點的映射位置應為之前沒出現的點，合並操作的時候也是要這樣操作映射數組！
所以AC代碼：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;

const int maxn=100000+10;
int f[maxn*10],a[maxn];

int find(int x)
{
    if(x!=f[x])
        f[x]=find(f[x]);
    return f[x];
}

void Union(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)return ;
    f[x]=y;
}

set<int>s;

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.cpp","r",stdin);
    freopen("out.cpp","w",stdout);
    #endif // ONLINE_JUDGE
    int n,m;
    char str[2];
    int cas=1;
    while(scanf("%d %d",&n,&m)==2&&(n+m))
    {
        s.clear();
        for(int i=0;i<=n;i++)
        {
            f[i]=i;
            a[i]=i;
        }
        int num=n;
        while(m--)
        {
            scanf("%s",str);
            if(str[0]==‘S‘)
            {
                int k;
                scanf("%d",&k);
                a[k]=num++;
                f[a[k]]=a[k];
            }
            else
            {
                int x,y;
                scanf("%d %d",&x,&y);
                Union(a[x],a[y]);
            }
        }

        for(int i=0;i<n;i++)
        {
            s.insert(find(a[i]));
        }
        printf("Case #%d: %d\n",cas++,s.size());
    }
    return 0;
}
 

HDU 2473 Junk-Mail Filter (並查集的刪除操作)