【并查集】poj 1611 The Suspects

The Suspects

Time Limit: 1000MS
Memory Limit: 20000KTotal Submissions: 39522
Accepted: 19129

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
题目大意：
有一个学校，有N个学生，编号为0-N-1，现在0号学生感染了非典，凡是和0在一个社团的人就会感染，
并且这些人如果还参加了别的社团，他所在的社团照样全部感染，求感染的人数。
解题思路：
并查集的变种，实质就是求0所在的强连通图的结点数目。
这道题纠结在数据的输入上，他只是告诉你哪些学生是同一个社团的。
这就需要处理一下，我的想法是：如果这个社团有num个孩子，new出一个大小为num的数组，
第一个孩子不处理，从第二个孩子起，和上个孩子合并，这样就完成了他们的合并。
///AC代码
#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define maxn 30010#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int a[maxn];int pre[maxn];int findd(int x)                           //查找根节点{    int r = x;    while ( pre[r] != r )                //返回根节点 r    {        r = pre[r];    }    int i = x, j ;    while ( i != r )                      //路径压缩    {        j = pre[i]; // 在改变上级之前用临时变量j记录下他的值        pre[ i ] = r ; //把上级改为根节点        i = j;    }    return r ;}void join(int x, int y)                    //判断x y是否连通，//如果已经连通，就不用管了 //如果不连通，就把它们所在的连通分支合并起,{    int fx = findd(x), fy = findd(y);    if (fx != fy)    {        pre[fx] = fy;    }    /*    else                                    ///同父节点，成环    {        flag = 0;    }    */}int main(){    int n, m;    while (scanf("%d%d", &n, &m) && n + m)    {        for (int i = 0; i < n; i++)        {            pre[i] = i;        }        int countt, ans = 0;        for (int i = 0; i < m; i++)        {            scanf("%d", &countt);            if (countt >= 1)            {                scanf("%d", &a[0]);                for (int j = 1; j < countt; j++)                {                    scanf("%d", &a[j]);                    join(a[0], a[j]);                }            }        }        for (int i = 0; i < n; i++)        {            if (findd(i) == pre[0])            {                ans++;            }        }        printf("%d\n", ans);    }    return 0;}