The Suspects

The Suspects

Time Limit: 1000MS
Memory Limit: 20000KTotal Submissions: 23963
Accepted: 11697

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 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

Source

Asia Kaohsiung 2003

1.并查集的变形；

思路：

1.这个题目中，每个小组是联通的，而且个人可以在不同的小组，所以考虑使用并查集；

2.将每个小组的在输入的时候遍历一遍，确定其中的最小根节点，因为其连通性，故将其中每一个值得根节点均置为该组的最小根节点；

这样遍历到最后的话，所有的点的根节点就都是最小的；

3.遍历出想过点，由于点没有连续性，所以在输入的时候顺便将mark数组标记下来；

4.统计；

AC代码如下：

#include <stdio.h>#include <string.h>#include <math.h>int pre[101000];int mark[100100];int find(int x){    int r=x;    while(r!=pre[r])    {        r=pre[r];    }    return r;}int main(){    int n,m,k;    int s[10010];    while(~scanf("%d%d",&n,&m)&&(n||m))    {        for(int i=1; i<=n; i++)        {            pre[i]=i;            mark[i]=0;        }        while(m--)        {            scanf("%d",&k);            int min1=999999;            for(int i=1; i<=k; i++)            {                scanf("%d",&s[i]);                mark[s[i]]=1;                if(find(s[i])<min1)                {                    min1=find(s[i]);                }            }            for(int i=1; i<=k; i++)            {                pre[find(s[i])]=min1;            }        }        mark[0]=1;        int sum=0;        for(int i=1; i<n; i++)        {            if(mark[i])            {                if(find(i)==0)                {                    sum++;                }            }        }        printf("%d\n",sum+1);    }    return 0;}