NO IMAGE

【題目】

題目傳送門Junk-Mail Filter

 

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

 

【分析】

題目大意:有多組資料,每組資料有 n 的點,m 個操作,以下兩種操作:

1、M X Y,將 X 和 Y 合併(放入同一個集合中)
2、S X,將 X 刪除

問題就是進過 m 次操作之後有多少個集合

這道題的話,操作 M 很好做,用並查集就行了,但是刪除不太好做,聽說並查集中刪點的程式碼要寫300多行

一個簡單的做法是對於每個要刪除的點 x,新建一個節點 y,然後將 x 指向 y,以後要用 x 的時候用 y 就可以了

emmm……是不是很簡單,接下來看程式碼吧

 

【程式碼】

這裡並查集是用路徑壓縮 按秩合併實現的

#include<cstdio>
#include<cstring>
#include<algorithm>
#define M 1000000 10
using namespace std;
int father[M],n,m;
int id[M],sum[M];
void init()
{
for(int i=0;i<M;i  )
{
father[i]=i;
id[i]=i;
sum[i]=1;
}
}
int find(int x)
{
if(father[x]!=x)
father[x]=find(father[x]);
return father[x];
}
void merge(int x,int y)
{
x=find(x);
y=find(y);
if(x==y) return;
if(sum[x]<sum[y])
swap(x,y);
father[y]=x;
sum[x] =sum[y];
sum[y]=0;
}
int main()
{
int n,m,i,x,y,ans,num=0;
char c;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(!n&&!m)
break;
ans=0;
num  ;
init();
for(i=1;i<=m;  i)
{
c=getchar();
c=getchar(); 
if(c=='M')
{
scanf("%d%d",&x,&y);
merge(id[x],id[y]);
}
if(c=='S')
{
scanf("%d",&x);
sum[find(id[x])]--;
id[x]=n  ;
}
}
for(i=0;i<n;i  )
if(sum[i])
ans  ;
printf("Case #%d: %d\n",num,ans);
}
return 0;
}