POJ 1611 :The Suspects(并查集)

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 21427		Accepted: 10375

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

题意：n，表示学生数目， m，表示社团数目。 每一个学生有一个编号。 0到N-1。 编号为0的学生是全部学生中最初的唯一感染者， 接下来由m个社团， 在每行输入一个k

，k表示社团的总人数， 接着是社团中k个成员的编号。

。 当输入n=0且m=0时，输入结束。

。

输出被感染的总人数。

。

解题思路：

本题编号为0的人已经被感染了，所以被感染的人数至少为1

运用并查集划分集合的同一时候记录每个集合 的元素个数。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, n) for(int i=1; i<=n; i++)
#define f3(i, n) for(int i=n; i>=1; i--)
#define f4(i, n) for(int i=2; i<=n; i++)
#define M 30050

int n;
int p[M];
int r[M];
int a[M];

void start()
{
    for(int i=0; i<=n-1; i++)
    {
        p[i]=i;  //初始化并查集
        r[i]=1; //初始化时，每一个元素作为一个集合，其元素为1
    }
}

int find(int x)   //并查集的find
{
    return p[x] == x ? x : p[x] = find ( p[x] );
}

void Kruskal(int x, int y)
{
    int xx = find(x);
    int yy = find(y);   //找出当前两个端点所在集合的编号

    //假设在不同集合就合并
    if(xx!=yy)
    {
        r[xx] += r[yy];  //合并时更改集合元素的总数
        p[yy] = xx;
    }
}

int main()
{
    int m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0 && m==0)
            break;
        start();
        while( m-- )
        {
            int t;
            scanf("%d", &t);
            f1(i, t)
            {
                scanf("%d", &a[i]);
                if(i!=0)
                    Kruskal(a[i-1], a[i]);
            }
        }
        printf("%d\n", r[find(0)]);//找到包括0元素的集合的节点。并输出其记录的节点数目
    }
    return 0;
}