POJ 1611 The Suspects

題目：

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 22216		Accepted: 10804

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

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解題思路：

並查集的簡單應用。檢查有多少個元素和元素0在同一個集合之中。我們可以在進行並操作時，將小的點設爲父親節點，這樣的話，所有與0在一個集合中的元素的根節點都爲0，我們找出有多少個點的根節點爲你0即可。

代碼：

#include <cstdio>
#include <cstring>

const int MAXN = 3e4 + 10;
int set[MAXN];

int find(int p)
{
    if(set[p] < 0) return p;
    return set[p] = find(set[p]);
}

void join(int p, int q)
{
    p = find(p), q = find(q);
    if(p < q) set[q] = p;
    if(p > q) set[p] = q;
}

int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m))
    {
        if(0==n && 0==m) break;
        memset(set, -1, sizeof(set));
        while(m--)
        {
            int num, a;;
            scanf("%d", &num);
            scanf("%d", &a); num--;
            while(num--)
            {
                int b;
                scanf("%d", &b);
                join(a, b);
            }
        }
        int cnt = 1;
        for(int i = 1; i < n; ++i)
        {
            if(0 == find(i)) cnt++;
        }
        printf("%d\n", cnt);
    }
    return 0;
}