1.19--J

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

这题思路是参考了别人的解法的，算法是并查集。
题解：
0是患者，与0在一个集合里面的人数就是答案，把每一行中的编号看成是在一个集合里面，找到每个集合里的每个元素的根节点，若根节点不同就连在一起，然后会形成几颗类似树一样形式的集合，为了减少路径，用了rank数组来表示每个根节点有多少层树杈，树杈少的就附在树杈多的上面。

以下是AC代码：

#include<iostream>
using namespace std;
int parents[30005], ranks[30005];          //parents数组用来记录根节点，ranks数组用来记录有多少层树杈
int find_root(int x)              //找根节点函数
{
	while (parents[x] != -1)              //表示有根节点，然后一层层往上面寻找，直到找出最终根节点
		x = parents[x];
	 return x;             //假设没有根节点，即parents[x]=-1，都是直接返回x的数值
}
void union_root(int a, int b)            //链接两个根节点函数
{
	a = find_root(a);               //每个值都需要找根节点
	b = find_root(b);
	if (a != b)                     //根节点不同表示不在一个集合里面，这时就需要连接两个根节点
		if (ranks[a] > ranks[b])                       //比较谁的根节点树杈层数多，层数少的附在层数多的上面
			parents[b] = a;
		else if (ranks[a] < ranks[b])
			parents[a] = b;
		else if (ranks[a] == ranks[b])
		{
			parents[b] = a;
			ranks[a]++;                //树杈层数相同，a附在b上，或者b附在a上都可以，但是无论怎么附着，该最终根节点都会加一
		}
}
int main()
{
	int n, m;
	for (int i = 0; i < 30005; i++)
		parents[i] = -1, ranks[i] = 0;
	while (cin >> n >> m)
	{
		int t = 1;                   //t从1开始是因为0是患者
		if (n == 0 && m == 0)
			break;
		while (m--)
		{
			int k,* a;
			cin >> k;
			a = new int[k];
			for (int i = 0; i < k; i++)
				cin >> a[i];
			for (int i = 0; i < k-1; i++)
				union_root(a[i], a[i + 1]);
		}
		for (int i = 1; i <= n; i++)
			if (find_root(i) == find_root(0))               //只要每个编号的根节点与0相同，就意味着在一个集合里面，这时可以计算出人数
				t++;
		cout << t << endl;
		for (int i = 0; i < 30005; i++)
			parents[i] = -1, ranks[i] = 0;
	}
}