Codeforces 170C Learning Languages (并查集求连通分支)

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本文介绍了如何应用并查集解决Codeforces上的一个编程挑战，该挑战涉及找到公司中员工之间的最小花费，以便任何员工都能通过翻译与其他人通信。题目描述了公司员工和他们已知的语言，目标是确定使所有员工都能通信所需的最低学习成本。

The "BerCorp" company has got n employees. These employees can use m approved official languages for the formal correspondence. The languages are numbered with integers from 1 to m. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.

Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 100) — the number of employees and the number of languages.

Then n lines follow — each employee's language list. At the beginning of the i-th line is integer k_i (0 ≤ k_i ≤ m) — the number of languages the i-th employee knows. Next, the i-th line contains k_i integers — a_ij (1 ≤ a_ij ≤ m) — the identifiers of languages the i-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.

The numbers in the lines are separated by single spaces.

Output

Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).

Sample test(s)

input
5 5
1 2
2 2 3
2 3 4
2 4 5
1 5

output
0

input
8 7
0
3 1 2 3
1 1
2 5 4
2 6 7
1 3
2 7 4
1 1

output
2

input
2 2
1 2
0

output
1

Note

In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.

In the third sample employee 2 must learn language 2.

用并查集来求连通分支，也就是没有共同语言的团体数，当然这里需要注意特判，当所有人都没有共同语言的时候。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=101;
const int maxm=101;
int cnt[maxm];
int peo[maxm][maxn];
int pre[maxn],n,m;
void  init()
{
    int i;
    for(i=1; i<=n; i++)
        pre[i] = i;
}

int find(int x)
{
    int r=x;
    while(r!=pre[r])
        r=pre[r];
    int i=x,j;
    while(i!=r)
    {
        j=pre[i];
        pre[i]=r;
        i=j;
    }
    return r;
}

void unon(int a,int b)
{
    int at = find(a);
    int bt = find(b);
    if(at==bt) return;
    else
    {
        pre[at] = pre[bt];
    }
}
int main()
{
    int k,v;
    while(~scanf("%d%d",&n,&m))
    {
        memset(peo,0,sizeof(peo));
        memset(cnt,0,sizeof(cnt));
        init();
        bool f=false;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&k);
            if(k) f=true;
            while(k--)
            {
                scanf("%d",&v);
                peo[v][cnt[v]++]=i;
            }
        }
        for(int i=1; i<=m; i++)
            for(int j=1,u=peo[i][0]; j<cnt[i]; j++)
                unon(peo[i][j],u);
        int res=-1;
        for(int i=1; i<=n; i++)
            if(pre[i]==i) res++;
        if(f) printf("%d\n",res);
        else printf("%d\n",res+1);
    }
    return 0;
}