POJ1611 The Suspects 种类并查集

1.题目描述：

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 36019		Accepted: 17519

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

2.题意概述：

n个人，m组关系，每组关系中有k个人，与0相关的均为嫌疑人，求有多少嫌疑人

3.解题思路：

并查集加一个秩来统计一下人数，然后查询时候直接查询0的父亲的人数即可

4.AC代码：

#include <stdio.h>
#include <iostream>
#define maxn 30300
using namespace std;

int pa[maxn];        //存储有向图的边
int sn[maxn];
void init()     //初始化	该函数可以根据具体情况保存和初始化需要的内容  
{
	for (int i = 0; i < maxn; i++)
	{
		pa[i] = i;
		sn[i] = 1;
	}
}

int findset(int a)	//不带路劲压缩  
{
	while (pa[a] != a)
		a = pa[a];
	return a;
}

void unionset(int a, int b)      //集合合并  
{
	int a1 = findset(a);
	int b1 = findset(b);
	if (a1 != b1)		//这个判定条件可选，主要是为了防止findset路径压缩的时候出现死循环
	{
		pa[a1] = b1;		//如果存的是有向图，并且做题时集合中元素的顺序很重要，不能忽略，那么这里应该用"pa[a] = b;" 
		sn[b1] += sn[a1];
	}
}

int main()
{
	int num, gp;
	while (scanf("%d%d", &num, &gp) != EOF && num)
	{
		init();
		while (gp--)
		{
			int n_num;
			scanf("%d", &n_num);
			int * student = new int[n_num];
			for (int j = 0; j < n_num; j++)
			{
				scanf("%d", &student[j]);
				if (j)
					unionset(student[j], student[j - 1]);
			}
			delete[] student;
		}
		printf("%d\n", sn[findset(0)]);
	}
	return 0;
}