題目描述：
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.
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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.
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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.
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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
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Sample Output
?
Source
?
Recommend
lcy
分析：
有于有可能要刪掉以前的根節點。所以要保存它的信息。
a[i] 表是原來編號為i的字母的位置，例如原來是a[i]=i;
現在刪掉i后，假如 i 原來是一個集合的根節點。那么，只對b[i]--;
即這個集合的個數 減一。而這個i? 的位置則移動到數組最后面 a[sum]中。
此時a[i]=sum;完成i的轉移。
因而當findroot(i)時應該變成 find(a[i])，
相當于尋址。MY ACCEPTED CODE：
/**//*
******************************************************
HDU?2473?Junk-Mail?Filter?(2008Regional杭州現場賽題目)
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;????
}