题意

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).

输入格式

• 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.

输出格式

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).

样例

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

有两种人：一种是一种语言都不会的，一种是会至少一种语言的。
会语言的人用并查集记录，只要是会同一种语言的就合并。

A: 一种语言都不会的人。
B: 另外不同圈子之间要交流，也就是还需要培训(圈子数 - 1)个人。
需要培训的总人数就是 A+B 。

代码

#include 

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;
}