A - The Suspects C语言

A - The Suspects
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

注:这题很明显考查”并查集”的问题.但问题就是后面m行数据很难处理,我刚做的时候也在考虑输入数据的问题, 总不能定义个数据将所有数记录吧,很明显实现不了,若用for,那上一组的关系结果怎么来记录便是一个问题, 于是我们不妨定义一个num[i]数组,给他赋初值为1,这样我们再for一下,每一组数都建立联系,用num[这个是这组数的根节点]记录每组有多少人,从0开始遍历查询0的根节点,num[根节点]就是所求.
代码如下:

#include <stdio.h> 

int num[30005] , bin[30005] , a[30005];
int find(int aa)
{
    return aa==bin[aa] ? aa : find(bin[aa]); //得到aa的根节点 
}
int jjj(int x , int y)
{
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
    {
        bin[xx]=yy;                 
        num[yy]+=num[xx];           //以yy为根节点的人数，后面只要查找0的根结点就知道有多少 
    }                               // 人了 
    return 0;                   
}

int main()
{
    int n , m , b;
    while(~scanf("%d%d", &n , &m))
    {
        if(n==0 && m==0)
            return 0;
        for(int i=0 ; i<n ; i++)
        {
            bin[i]=i;
            num[i]=1;   
        }
        while(m--)
        {
            scanf("%d", &b);
            for(int i=0 ; i<b ; i++)
            {
                scanf("%d", &a[i]);
            }
            for(int i=0 ; i<b-1 ; i++)
            {
                jjj(a[i] , a[i+1]);
            }
        }
        int k=find(0);          //查找0的根节点 
        printf("%d\n", num[k]);

    }
    return 0;
}