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.

输入描述

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 A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

输出描述

For each case, output the number of suspects in one line.

输入例子

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

输出例子

4
1
1

参考答案

#include <stdio.h>
#include <iostream>
using namespace std;

const int MAXN = 30001; /*结点数目上线*/
int pa[MAXN];    /*p[x]表示x的父节点*/
int rank[MAXN];    /*rank[x]是x的高度的一个上界*/
int num[MAXN];/*num[]存储该集合中元素个数，并在集合合并时更新num[]即可*/

void make_set(int x)
{/*创建一个单元集*/
    pa[x] = x;
    rank[x] = 0;
    num[x] = 1;
}

int find_set(int x)
{/*带路径压缩的查找*/
    /*保存待查找的数*/
    int r = x, temp;
    /*找到根节点*/
    while(pa[r] != r) r = pa[r];
    while(x != r)
    {
        temp = pa[x];
        pa[x] = r;
        x = temp;
    }
    return x;
    //if(x != pa[x]) //注释掉的其实也是可以的，不过不想用递归来做啦
    //    pa[x] = find_set(pa[x]);
    //return pa[x];
}

/*按秩合并x，y所在的集合*/
void union_set(int x, int y)
{
    x = find_set(x);
    y = find_set(y);
    if(x == y)return ;
    if(rank[x] > rank[y])/*让rank比较高的作为父结点*/
    {
        pa[y] = x;
        num[x] += num[y];
    }
    else 
    {
        pa[x] = y;
        if(rank[x] == rank[y])
            rank[y]++;
        num[y] += num[x];
    }
}

int main()
{
    int n, m, x, y, i, t, j;
    while(scanf("%d%d", &n, &m))
    {
        if(m==n && n == 0) break;
        if(m == 0)
        {
            cout << "1\n"; continue;
        }
        for(i = 0; i < n; i++)
            make_set(i);
        for(i = 0; i < m; i++)
        {
            scanf("%d", &t);
            scanf("%d", &x);
            for(j = 1; j < t; j++){
                scanf("%d", &y);
                union_set(x, y);
                x = y;
            }
        }
        x = find_set(0);/*找到0所在的树的树根*/
        cout << num[x] << endl;
    }
    return 0;
}