The Suspects

Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 33378 Accepted: 16192
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

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这一题用自己的方法比较麻烦，就是先分集合，再搜索元素个数，逐个加一，容易错，看了一下别人的题解，思维很巧妙，用rank控制和0 同一集合的元素的个数，连接成一条直线，在合并时直接相加，最后求长度。不失为一种很好的办法。就是适用性不强。

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#include<cstdio>
#include<iostream>

using namespace std;

int const MAX_N = 30010;  

int par[MAX_N];           
int rankk[MAX_N];       
int student[505];

void init( int n )
{
    for( int i=0; i<n; i++ )
    {
        par[i]=i;
        rankk[i]=1; //全部先设为1，方便后面求长度
    }
} 

int find( int x )
{
    if( par[x] == x )
    return x;
    else return find(par[x]);  
}

void unite( int x, int y )
{
    x = find(x);
    y = find(y);
    if( x == y ) return;
    if( rankk[x] < rankk[y] )
    {
        par[x] = y;
        rankk[y] = rankk[x]+rankk[y];//关键
    }else
    {
        par[y] = x;
        rankk[x] = rankk[x]+rankk[y];
    }
}

int main()
{
    int n,m,k,a,b;
    int first,next;
    while( ~scanf("%d%d",&n,&m),n||m )
    {
        if( m == 0 ){
            printf("1n");
            continue;
        }
        init(n);
        for( int i=0; i<m; i++ )
        {
            scanf("%d",&k);
            scanf("%d",&first);
            for( int j=1; j<k; j++ )
            {
                scanf("%d",&next);
                unite(first,next);
            }
        }
    // 废置代码
    //  int count = 0;
    //  for( int i=1; i<=n; i++ )
    //  {
    //      if( find(0) == find(i) )
    //      count++;
    //  }
        printf("%dn",rankk[par[0]]);
    }
}

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