POJ 1611 The Suspects(并查集)

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 25290 Accepted: 12396

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

解题思路:

不用多想，就是并查集，我们只需先合并团体，把每行的第一个元素作为根(union函数)，然后根再与0判断一下是否在一个团体内(find函数)，然后我们从结点０开始遍历一遍，如果和0在一个团体内就cnt++,最后的cnt便是结果。

AC代码:

#include<iostream>#include<cstdio>using namespace std;const int maxn = 30001;const int maxn1 = 501;int a[maxn][maxn1];int father[maxn];int find(int x){    if(x != father[x])    {        father[x] = find(father[x]);    }    return father[x];}void Union(int a,int b){    int fx = find(father[a]);    int fy = find(father[b]);    if(fx != fy)    {        father[fy] = fx;    }}int main(){    int m,n;    int i,j;    int h;    int cnt;    //freopen("111","r",stdin);    while(cin>>m>>n && m + n)    {        cnt = 0;        for(i=0;i<m;i++)        {            father[i] = i;        }        for(i=0;i<n;i++)        {            cin>>h;            for(j=0;j<h;j++)            {                cin>>a[i][j];                Union(a[i][0],a[i][j]);            }        }        for(i=0;i<m;i++)        {            if(find(i) == find(0))            {                cnt++;            }        }        cout<<cnt<<endl;    }    return 0;}