题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2473

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.

题目分析：这个题目是一个典型的考查并查集的题目，主要思路是：遇到'M'则合并集合，遇到S则将 S 后面的 X 结点变成独立的结点！主要的是把父亲结点为X 的结点，修改为父亲结点为自己！！并且把X的父亲结点置为X！

例如：

所以我用一个replace数组来标志所有的结点，来处理上述问题。

代码如下：

View Code

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<memory.h>
 4 int set[1100002] , flag[1100002] ;
 5 int replace[100002] ;
 6 int inf = 0 ;
 7 int find_index(int x)  //查 根
 8 {
 9     int r = x , a = x , b ;
10     while(set[r] != r)
11     {
12         r = set[r] ;
13     }
14     while(a != r)
15     {
16         b = set[a] ;
17         set[a] = b ;
18         a = b ;
19     }
20     return r ;
21 }
22
23 void merg(int x , int y)   //并 集合
24 {
25     int fx , fy ;
26     fx = find_index(x) ;
27     fy = find_index(y) ;
28     if(fx > fy)
29         set[fx] = fy ;
30     else
31         set[fy] = fx ;
32 }
33
34 void del(int x)       //分离结点
35 {
36     replace[x] = inf ;
37     set[inf] = inf ;
38     inf++ ;
39 }
40 int main(void)
41 {
42     int n , m , lines = 1 ;
43     while(scanf("%d%d" , &n , &m) != EOF && !(m == 0 && n == 0))
44     {
45         int i , j , x , y , cnt = 0 ;
46         char ch ;
47         for(i = 0 ; i < n ; ++i)
48             set[i] = replace[i] = i ;
49         inf = i ;
50         for(i = 0 ; i < m ; ++i)
51         {
52             getchar() ;
53             scanf("%c" , &ch) ;
54             if(ch == 'M')
55             {
56                 scanf("%d%d" , &x , &y) ;
57                 merg(replace[x] , replace[y]) ;
58             }
59             else
60             {
61                 scanf("%d" , &x) ;
62                 del(x) ;
63             }
64         }
65         memset(flag , 0 , sizeof(flag)) ;
66         for(i = 0 ; i < n ; ++i)
67         {
68             x = find_index(replace[i]) ;
69             if(!flag[x])
70             {
71                 cnt++ ;
72                 flag[x] = 1 ;
73             }
74         }
75         printf("Case #%d: %d\n" , lines++ , cnt) ;
76     }
77     return 0 ;
78 }