POJ1611_ The Suspects

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 22366		Accepted: 10875

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 4 2 1 2 5 10 13 11 12 14 2 0 1 2 99 2 200 2 1 5 5 1 2 3 4 5 1 0 0 0

Sample Output

4 1 1

Source

​ ​Asia Kaohsiung 2003​​

 

The main idea of the topic ：

n Students belong to m Groups ,(0 < n <= 30000 , 0 <= m <= 500) A student can belong to multiple groups . A student is suspected to be ill , Then the whole group it belongs to is suspected to be sick . It is known that 0 Student No. 1 is suspected to be ill , And which students make up each group , Ask how many students are suspected to be ill .

Ideas ： The most basic application of union search set . Put all suspicious things together . among , Use a total Array ,total[find(a)] yes a Where
Number of people gathered . The final output 0 The total number of infections in the set of node , That is to say 0 The total number of infections in the root node of node .

Code ：

Log in and copy

      
      

       
       # include<stdio.h>
       
       

       
       # include<string.h>
       
       

       
       const 
       
       int 
       
       MAXN 
       
       = 
       
       30010;
       
       

       
       int 
       
       father[
       
       MAXN],
       
       total[
       
       MAXN];
       
       

       
       

       
       int 
       
       find(
       
       int 
       
       a)
       
       // obtain a The root of the , And put a The parent node of is changed to root 
       
       
{
       
       
    
       
       if(
       
       a 
       
       != 
       
       father[
       
       a])
       
       
        
       
       father[
       
       a] 
       
       = 
       
       find(
       
       father[
       
       a]);
       
       
    
       
       return 
       
       father[
       
       a];
       
       
}
       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       n, 
       
       m, 
       
       k, 
       
       a, 
       
       b;
       
       
    
       
       while(
       
       scanf(
       
       "%d %d", 
       
       &
       
       n, 
       
       &
       
       m) 
       
       && (
       
       n
       
       ||
       
       m) )
       
       
    {
       
       
        
       
       memset(
       
       father,
       
       0,
       
       sizeof(
       
       father));
       
       
        
       
       memset(
       
       total,
       
       0,
       
       sizeof(
       
       total));
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       father[
       
       i] 
       
       = 
       
       i;
       
       
            
       
       total[
       
       i] 
       
       = 
       
       1;
       
       
        }
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       m; 
       
       i
       
       ++)
       
       
        {
       
       
            
       
       scanf(
       
       "%d", 
       
       &
       
       k);
       
       
            
       
       scanf(
       
       "%d", 
       
       &
       
       a);
       
       
            
       
       for(
       
       int 
       
       j 
       
       = 
       
       1; 
       
       j 
       
       < 
       
       k; 
       
       j
       
       ++)
       
       
            {
       
       
                
       
       scanf(
       
       "%d", 
       
       &
       
       b);
       
       
                
       
       a 
       
       = 
       
       find(
       
       a);
       
       
                
       
       b 
       
       = 
       
       find(
       
       b);
       
       
                
       
       if(
       
       a
       
       !=
       
       b)
       
       
                {
       
       
                    
       
       father[
       
       b] 
       
       = 
       
       a;
       
       
                    
       
       total[
       
       a] 
       
       += 
       
       total[
       
       b];
       
       
                }
       
       
                
       
       
            }
       
       
        }
       
       

       
       
        
       
       printf(
       
       "%d\n",
       
       total[
       
       find(
       
       0)]);
       
       // What must be output is 0 Total number of root nodes , Because path compression may not be done yet .
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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