POJ - 1611 The Suspects （并查集）

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
问题链接：http://poj.org/problem?id=1611
问题简述：意思是先输入n代表n个学生，然后有m个组，下面输入每个学生的序号（从1到n-1），0号学生是病原体。判断有多少名学生有可能感染。
问题分析：并查集水题…套模板可以直接A。。。（并查集主要是写find函数跟union函数吧，把两个函数写熟了就OK了，以后并查集估计应该是和别的题目结合的了吧…）
AC通过的C++语言程序如下：

#include<iostream>
#include <algorithm>
#include<iostream>
#include<string>
#include<stdio.h>
#include <algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 30005;
int n, m;
int parent[maxn], num[maxn], an[maxn];
void ini()
{
	for (int i = 0; i < n; i++)
	{
		parent[i] = i;
		num[i] = 1;
	}
}
int find(int x)
{
	if (x == parent[x])
	{
		return x;
	}
	else return parent[x] = find(parent[x]);
}
void unionn(int x, int y)
{
	int root_x = find(x);
	int root_y = find(y);
	if (root_x != root_y)
	{
		parent[root_x] = root_y;
		num[root_y] = num[root_y] + num[root_x];
	}
	return;
}
int main()
{
	while (cin >> n >> m)
	{
		if (n == 0 && m == 0)return 0;
		ini();
		for (int i = 0; i < m; i++)
		{
			int t;
			cin >> t;
			for (int i = 0; i < t; i++)
			{
				cin >> an[i];
			}
			for (int i = 0; i < t - 1; i++)
			{
				unionn(an[i], an[i + 1]);
			}
		}
		int res = find(0);
		cout << num[res] << endl;
	}
	return 0;
}