HDOJ 题目2473 Junk-Mail Filter（并查集，删点）

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本文介绍了一种通过分析垃圾邮件样本并提取共同特征来识别垃圾邮件的方法。该方法使用了两个主要步骤：首先从邮件中提取特征，然后利用这些特征进行匹配以判断邮件是否为垃圾邮件。文中还详细描述了一个用于跟踪不同邮件间关系的数据结构及其操作。

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Junk-Mail Filter

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6406 Accepted Submission(s): 2031

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 ≤ 10 ⁵ , 1 ≤ M ≤ 10 ⁶), 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

Sample Output

  
   Case #1: 3
Case #2: 2

Source

2008 Asia Regional Hangzhou

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ac代码

#include<stdio.h>
#include<string.h>
int pre[2200000],flag[2200000];
int n,m,id;
int find(int x)
{
	if(x==pre[x])
		return x;
	else
		return pre[x]=find(pre[x]);
}
void join(int a,int b)
{
	int fa,fb;
	fa=find(a);
	fb=find(b);
	if(fa!=fb)
	{
		pre[fb]=fa;
	}
}
void del(int a)
{
	pre[a]=id++;
}
int main()
{
	//int n,m;
	int cot=0;
	while(scanf("%d%d",&n,&m)!=EOF,n||m)
	{
		int i;
		id=n+n;
		for(i=0;i<n;i++)
			pre[i]=i+n;
		for(i=n;i<=n+n+m;i++)
			pre[i]=i;
		for(i=0;i<m;i++)
		{
			char c;
			getchar();
			scanf("%c",&c);
			if(c=='M')
			{
				int a,b;
				scanf("%d%d",&a,&b);
				join(a,b);
			}
			else
			{
				int a;
				scanf("%d",&a);
				del(a);
			}
		}
		int ans=0;
		memset(flag,0,sizeof(flag));
		for(i=0;i<n;i++)
		{
			int x=find(i);
			if(flag[x]==0)
			{
				ans++;
				flag[x]=1;
			}
		}
		printf("Case #%d: %d\n",++cot,ans);
	}
}