POJ 1611 The Suspects 并查集（代码带启发式合并）

The Suspects
Time Limit: 1000MS
 
Memory Limit: 20000K
Total Submissions: 23213
 
Accepted: 11268
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 
  
 
  
 
  
 
  
并查集模板题，带启发式合并、路径压缩
 
  
把同一社团中的所有成员并成一个集合，加入多个社团的人会使这些社团合为一个集合
 
  
这样输出0所在集合的元素个数即可。
 
  
因为要记录元素个数，顺便将代码加入启发式合并
 
  

 
  
#include
#include

const int MAX = 30000 + 100;
int fa[MAX];

int Find(int x){
    return fa[x] < 0 ? x : fa[x] = Find(fa[x]);///根的fa[]值小于0
}

void Union(int x,int y){
    int a = Find(x),b = Find(y);
    if(a == b)
        return;
    ///启发式合并，判断哪个集合的元素多，将元素少的合并到元素多的集合上去
    ///根节点的fa[]值的绝对值即为集合中元素个数
    if(fa[a] > fa[b]){///注意值为负，大的绝对值小
        fa[b] += fa[a];
        fa[a] = b;
    } else {
        fa[a] += fa[b];
        fa[b] = a;
    }
}

int main(void) {
    int m,n;
    while(~scanf("%d%d",&n,&m) && (m || n)){
        memset(fa,-1,sizeof(int) * n);///fa[]数组全部初始化为-1，即元素个数为1的根
        for(int i = 0;i < m;++i){
            int k;
            scanf("%d",&k);
            if(k > 0){
                int x,y;
                scanf("%d",&x);
                for(int i = 1;i < k;++i){
                    scanf("%d",&y);
                    Union(x,y);
                }
            }
        }
        printf("%d\n",-fa[Find(0)]);
    }
    return 0;
}