一、题目：

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 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.

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 1

Inputcopy	Outputcopy
5 5 1 2 2 2 3 2 3 4 2 4 5 1 5	0

Sample 2

Inputcopy	Outputcopy
8 7 0 3 1 2 3 1 1 2 5 4 2 6 7 1 3 2 7 4 1 1	2

Sample 3

Inputcopy	Outputcopy
2 2 1 2 0	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.

题意：

公司一共有N个人，一共官方有M种语言，如果A会语言1，2，B会语言2,3 C会语言3 4，则ABC三个人就可以相互交流了，即是一个集合。

公司为了让这N个人都能够相互交流，可以使得一些人学会一门语言，对应花费为1，问最少花费多少？

二、思路：

能够相互说话的元素合并成一个集合，则该集合的每个元素的根节点相同，不同集合的集合根节点不同，若有两个集合，只需要向其中一个集合加入另一结合中存在的语言即可合并，则x个集合，就花费x-1元，但x==n时，此时有n个集合，但每个人至少的要学一种语言，则花费n元。

三、代码：

#include<iostream>
using namespace std;
int parent[205];
int find(int x) {if (parent[x] == x) return x;return parent[x] = find(parent[x]);//状态压缩
}
void unit(int r, int y) {int r1 = find(r);int y1 = find(y);parent[y1] = r1;
}
int main()
{int n, m, l, k;int flag = 0;cin >> n >> m;int ans = 0;for (int i = 1; i <= n + m; i++) {parent[i] = i;//初始化}for (int i = 1; i <= n; i++) {cin >> l;if (l == 0) continue;elsefor (int j = 1; j <= l; j++) {cin >> k;unit(i, k + n);//一人为根节点flag = 1;}}for (int i = 1; i <= n; i++) {if (parent[i] == i)ans++;//parent[i] == i，遍历根节点数，表示集合数。}if (!flag)cout << n;//特判每人至少学一个语言。else cout << ans - 1;
}

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B - Learning Languages

一、题目：

二、思路：

三、代码：

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