Codeforces 278C. Learning Languages

Learning Languages Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

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Description

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 Input

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

Hint

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.

题意：公司有N个员工，每个人可能会M种语言，要想让所有人能直接或间接沟通，至少需要学多少人次。

解法：只需要用并差集找出已经形成集合的个数n，那么还需要n-1就能把整个合并成一个集合，然后加上一门

语言都不会的人数即可，需要注意磁体有特数据，所有人会的语言数都为0的情况。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
const int maxn=101;
int n,m,fa[maxn];
set<int> s[maxn];

int find(int x){
    return x==fa[x] ? x : fa[x]=find(fa[x]);
}

bool check(int i,int j){
    set<int>::iterator x=s[i].begin();
    while(x!=s[i].end()){
        int y=*x;
        if(s[j].find(y)!=s[j].end()) return true;
        x++;
        //cout << 1 << endl;
    }
    return false;
}

int main()
{
    scanf("%d %d",&n,&m);
    for(int i=0;i<=n;i++) fa[i]=i;
    int sum,x,ans=0;
    for(int i=0;i<n;i++){
        scanf("%d",&sum);
        if(sum==0) ans++;
        while(sum--){
            scanf("%d",&x);
            s[i].insert(x);
        }
    }
    if(ans==n) printf("%d\n",n);
    else{
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                if(check(i,j)){
                    int fi=find(i),fj=find(j);
                    if(fi!=fj) fa[fi]=fj;
                }
        int ant=0;
        for(int i=0;i<n;i++) if(fa[i]==i) ant++;
        printf("%d\n",ant-1);
    }
    return 0;
}