POJ-1611 The Suspects

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这是一个关于算法竞赛的题目，描述了在SARS疫情期间，为了防止病毒传播，需要将疑似病例隔离开来。题目给出了每个学生所属的小组，并要求一旦小组内出现疑似病例，整个小组都要被标记为疑似。提供的C++代码实现了一个并查集的解决方案，通过查找每个学生所属的集合根节点，来确定疑似病例的数量。

The Suspects

Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 71516 Accepted: 33724

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

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int p[30010];
int find(int x)
{
	if(p[x]!=x)
	return p[x]=find(p[x]);
	return x;
};

int main()
{
	int n,m;
	int a,b,c;
	
	while(cin>>n>>m)//n总人数 
	{
		if(n==0&&m==0)break;
		int ans=0;
		for(int i=0;i<=n-1;i++)p[i]=i;
		for(int i=1;i<=m;i++)
		{
			cin>>a;
		    cin>>b;//嫌疑人 
		    for(int i=2;i<=a;i++)
		    {
		    	cin>>c;
		    	p[find(c)]=find(b);
			}
		}
		for(int i=0;i<=n-1;i++)
		if(find(p[find(i)])==p[0])ans++;//"最初学生 0 在所有案例中都被确认为嫌疑人" 
		cout<<ans<<endl;
		
	}
	return 0;
}