1611:The Suspects

描述
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.
输入
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.
输出
For each case, output the number of suspects in one line.

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题目大意：0号学生患病，跟0号学生直接同组或间接同组的同学也是患病的嫌疑人。求嫌疑人个数。

思路：并查集

因为要求最后的嫌疑人个数，所以需要记录每棵树的成员数目。该数目由每棵树大树根来记录就好了。

最后输出的结果是0号节点所在的那棵树的数目。

树在合并时，需要考虑效率问题，小树合并至大树上，即大树的根节点成员数目更新。

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AC代码

#include <iostream>
#include <algorithm>

using namespace std;

int stu[30000]; //记录学生的根节点 stu[i]=i时，i为根节点
int num[30000]; //记录成员个数。如果i号节点为根节点，则num[i]为该棵树的成员数目

int FindParent(int i){
if( stu[i]!= i )
stu[i]=FindParent(stu[i]);
else
return i;
return stu[i]; //parent＝root
}

void UnionTree(int i, int j){
int ir,jr;
ir=FindParent(i);
jr=FindParent(j);
if(ir == jr)
return;
else{
if(num[ir]>num[jr]){
stu[jr] = ir;
num[ir]+=num[jr];
}
else{
stu[ir]=jr;
num[jr]+=num[ir];
}
}
}

int main(){
int m,n;
while(1){
cin >> n >> m;
if(n==0 && m==0)
break;
for(int i=0; i<n; i++){
height[i]=1;
num[i]=1;
stu[i]=i;
}
for(int i=0; i<m; i++){
int number,root, member;
cin >> number >> root;
for(int j=0; j<number-1; j++){
cin>>member;
UnionTree(root, member);
}
}

cout<<num[FindParent(0)]<<endl;
}
}