poj 1611 The Suspects_不见不散的结局是曲终人散_新浪博客

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 27759		Accepted: 13550

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：第一行输入两个树M,N；N代表下面的行数；M代表数的最大值，然后一次的每一行，第一哥数代表这一样数的个数；其中0为病原体，与病原体一行的数能被感染，最后问有多少个数被感染；

思路：并查集，用一个sum数组记录同一个根节点里面数的个数；最后输出0的根节点所包含的数的个数就是结果；

代码如下：

#include< stdio.h >

#include< stdlib.h >

//

//num[]存储节点所在集合元素的个数

//father[]存储节点的父节点


int sum[30001],father[30001];

//查找x元素所在的集合，返回根节点 

int find(int x)

{

    // 压缩路径，将路径上所有r的子孙都连接到r上

    if(x!=father[x])

        x=find(father[x]);

    return father[x];

}

void panduan(int a,int b)//合并a,b所在的集合 

{

    a=find(a);

    b=find(b);

    if(a==b)//如果两个元素在同一个集合则不需要合并

        return ;

    if(sum[a]<=sum[b])//将小集合合并到大集合中，更新集合个数

    {

        father[a]=b;

        sum[b]+=sum[a];

    }

    else

    {

        father[b]=a;

        sum[a]+=sum[b];

    }

}

int main()

{

    int t,a,b,n,m;

    int i,j;

    while(1)

    {

        scanf("%d%d",&n,&m);

        if(n+m==0) break;

        for(i=0; i < n; i++)

        {

            sum[i]=1;

            father[i]=i;

        }

        for(j=0; j < m; j++)

        {

            scanf("%d",&t);

            scanf("%d",&a);

            for(i=1; i < t; i++)

            {

                scanf("%d",&b);

                panduan(a,b);

            }