題意

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).

輸入格式

• 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 ki (0 ≤ ki ≤ m) — the number of languages the i-th employee knows. Next, the i-th line contains ki integers — aij (1 ≤ aij ≤ 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.

輸出格式

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).

樣例

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.

思路

有兩種人：一種是一種語言都不會的，一種是會至少一種語言的。
會語言的人用並查集記錄，只要是會同一種語言的就合併。

A: 一種語言都不會的人。
B: 另外不同圈子之間要交流，也就是還需要培訓(圈子數 - 1)個人。
需要培訓的總人數就是 A+B 。

程式碼

#include <iostream>

using namespace std;

const int N = 110;

int n, m, cnt, k;
int p[N], lang[N], a[N];

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

void join(int x, int y) {
    int fx = find(x);
    int fy = find(y);
    if (fx != fy) p[fx] = fy;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) p[i] = i;

    for (int i = 1; i <= n; i ++ ) {
        cin >> k;
        if (!k) {
            cnt ++ ;
            continue;
        }
        for (int j = 1; j <= k; j ++ ) {
            cin >> a[j];
            lang[a[j]] ++ ;
            if (j > 1) join(a[j - 1], a[j]);
        }
    }

    int pl = 0;
    for (int i = 1; i <= m; i ++ )
        if (lang[i] && p[i] == i)
                pl ++ ;

    pl = max(0, pl - 1);
    printf("%d\n", cnt + pl);

    return 0;
}