hdu 2473 Junk-Mail Filter(並查集+虛擬節點)

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 DescriptionRecognizing 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.

InputThere 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.

OutputFor 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 6M 0 1M 1 2M 1 3S 1M 1 2S 33 1M 1 20 0

Sample Output Case #1: 3Case #2: 2

分析：這道題是一道典型的並查集的題目，關鍵是節點從集合中刪除的s操作
這裡使用了節點數的的下標+n作為父節點，這個位置只是標記父節點，並沒其他含義，等於是虛擬節點。
這樣當刪除一個節點時只用從這個節點中拿出來讓其父節點重新放在一個虛擬位置，即下標從n+n開始向後找。
最後是尋找獨立特點的集合。將這些父節點放在一個長度為n的num的數組中，裡邊放置的每個節點對應的父節點的位置，
然後對這個數組排序，找出其中不同父節點的數目即集合的個數。


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