hdu 2473 Junk-Mail Filter(并查集+虚拟节点)

Junk-Mail Filter

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 169 Accepted Submission(s): 58

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：这道题是一道典型的并查集的题目，关键是节点从集合中删除的s操作
这里使用了节点数的的下标+n作为父节点，这个位置只是标记父节点，并没其他含义，等于是虚拟的节点。
这样当删除一个节点时只用从这个节点中拿出来让其父节点重新放在一个虚拟的位置，即下标从n+n开始向后找。
最后是查找独立特点的集合。将这些父节点放在一个长度为n的num的数组中，里边放置的每个节点对应的父节点的位置，
然后对这个数组排序，找出其中不同父节点的数目即集合的个数。


#include <iostream>
#include <stdio.h>
#include <algorithm>
#define MAXNUM1 100005
#define MAXNUM2 1000005

using namespace std;
int father[MAXNUM1*2+MAXNUM2],num[MAXNUM1];

int find_father(int a)
{
    if(father[a]==a)return a;
    else return father[a]=find_father(father[a]);
}
int main()
{
    int n,m,ca,x,y,total;
    char op[3];
    ca = 0;
    while(scanf("%d%d",&n,&m)!=EOF&&n)
    {
        for(int i=0;i<n;i++)father[i]=i+n;
        for(int i=n;i<=n+n+m;i++)father[i]=i;
        total = n+n;
        for(int t=0;t<m;t++)
        {
            scanf("%s",op);
            if(op[0]=='M')
            {
                scanf("%d%d",&x,&y);
                int rx = find_father(x);
                int ry = find_father(y);
                if(rx!=ry)father[rx]=ry;
            }else{
                scanf("%d",&x);
                father[x] = total++;
            }
        }
        for(int i = 0; i < n; i++)num[i] = find_father(i);
        sort(num,num+n);
        int ans=1;
        for(int i = 1; i < n;i++)
        if(num[i]!=num[i-1])ans++;
        printf("Case #%d: %d\n",++ca,ans);
    }
    return 0;
}