The Suspects
Time Limit: 1000MS   Memory Limit: 20000K
Total Submissions: 18859   Accepted: 9130

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
题目大意:第0号同学感染了SARS病毒,凡是与第0号同学一个社团的都为病毒感染嫌疑人,而且与这些嫌疑人一个社团的的人都为病毒感染嫌疑人,找出SARS病毒感染者嫌疑人数。
解题方法:并查集,每次统计每个集合中嫌疑人的个数,合并时将子节点的数目加到父节点。
#include 

typedef struct
{
    int nCount;
    int parent;
    int rank;
}UFSTree;

UFSTree Student[30005];

void MakeSet(int n)
{
    for (int i = 0; i <= n; i++)
    {
        Student[i].nCount = 1;
        Student[i].parent = i;
        Student[i].rank = 0;
    }
}

int FindSet(int x)
{
    if (x != Student[x].parent)
    {
        return FindSet(Student[x].parent);
    }
    else
    {
        return x;
    }
}

void UnionSet(int x, int y)
{
    x = FindSet(x);
    y = FindSet(y);
    if (Student[x].rank > Student[y].rank)
    {
        Student[y].parent = x;
        Student[x].nCount += Student[y].nCount;
    }
    else
    {
        Student[x].parent = y;
        Student[y].nCount += Student[x].nCount;
        if (Student[x].rank == Student[y].nCount)
        {
            Student[y].rank++;
        }
    }
}

int main()
{
    int n, m, k, x, y;
    while(scanf("%d%d", &n, &m) != EOF && (n != 0 || m != 0))
    {
        MakeSet(n);
        for (int i = 0; i < m; i++)
        {
            scanf("%d", &k);
            scanf("%d", &x);
            for (int j = 1; j < k; j++)
            {
                scanf("%d", &y);
                if (FindSet(x) != FindSet(y))
                {
                    UnionSet(x, y);
                }
            }
        }
        printf("%d\n", Student[FindSet(0)].nCount);
    }
    return 0;
}