Learning Languages （并查集）

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#算法 #并查集

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本文讨论了如何利用并查集算法解决公司中不同员工因语言限制导致的通信障碍问题。通过学习和整合员工已掌握的语言资源，最小化公司为确保所有员工间通信所需的费用。

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

Output

Input

Output

Input

2 2
1 2
0

Output

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.

三月的周赛，一直到现在才A。。。

新学了个算法，并查集。

解题思路：

对于每个学过相同语言的人，将其祖先合并成同一个，然后通过间接地联系，将两者之间可以沟通者的祖先也合并为同一个。

要考虑到特殊情况，如果所有人都没有学过语言，则需要n dollar。

算法：

并查集。

http://wenku.baidu.com/view/f55a0d26aaea998fcc220e8f.html

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f[105],lan[105],r[105];
int find(int x)
{
if (x!=f[x])
f[x]=find(f[x]);
return f[x];
}
void uni(int x, int y)
{
x=find(x);
y=find(y);
if (x==y) return;
f[x]=y;
}
int main ()
{
int n,m,i,j,yu,num;
memset(r,0,sizeof(r));
bool all=true;
cin>>n>>m;
for (i=1; i<=n; i++)
f[i]=i;
for (i=1; i<=n; i++)
{
scanf("%d",&num);
if (num) all=false;
while(num--)
{
scanf("%d",&yu);
if (r[yu]==0)
r[yu]=i;
else
uni(i,r[yu]);
}
}
if (all)
cout<<n<<endl;
else
{
int ans=0;
for (i=1; i<=n; i++)
if (f[i]==i)
ans++;
cout<<ans-1<<endl;
}

return 0;
}