poj 1611（并查集）（B）

The Suspects

Time Limit: 1000MS
Memory Limit: 20000KTotal Submissions: 38864
Accepted: 18832

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

题意；有n个人，编号为0~n-1，有m个团队，每个团队有k个人，k个人分别为....

如果一个他团队人有一个人感染，整个团队的人都感染

问如果先0号人感染，那么总共有多少人会被感染

思路：简单的并查集    只是多加了一个表示数量的数组  两个集合并在一起时  将其数量放在他们的根上

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define inf 1e9#define maxn 30000int n,m;int pre[maxn];int num[maxn];int a[maxn];int find(int x){int r=x;while(pre[r]!=r){r=pre[r];}int i=x,j;while(i!=r){j=pre[i];pre[i]=r;i=j;}return r;}void join(int i,int j){int fx=find(i),fy=find(j);if(fx!=fy){pre[fx]=fy;num[fy]+=num[fx];}}int main(){while(~scanf("%d%d",&n,&m)){if(m==0&&n==0)break;for(int i=0;i<n;i++)pre[i]=i,num[i]=1;int k;for(int j=1;j<=m;j++){scanf("%d",&k);for(int i=1;i<=k;i++)scanf("%d",&a[i]);for(int i=2;i<=k;i++)join(a[i-1],a[i]);}int z=find(0);printf("%d\n",num[z]);}}