Codeforces 278C Learning Languages【并查集】水题

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本文介绍了一种算法，用于计算使公司所有员工能通过一种或多种官方语言进行有效沟通所需的最小培训成本。通过使用并查集数据结构，该算法有效地处理了员工之间的语言互通性，并确定了最少的学习需求。

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).

Examples

Input

Output

Input

Output

Input

2 2
1 2
0

Output

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.

题目大意：

公司一共有N个人，一共官方有M种语言，其中已知每个人会的语言，如果两个人会同一种语言，那么两个人就可以相互交流，如果A会语言1，2，B会语言2,3 C会语言3 4，那么B可以给C翻译A说的话，说以这时候根据这个特性，那么ABC三个人就可以相互交流了。

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，问最少花费多少能够达到目的。

思路：

1、首先设定map【i】【j】，表示第i个人是否会第j门语言，那么我们接下来O（n^2*M）预处理出来哪些人可以相互交流，对应我们可以将这一堆人抽象合并成一个集合，那么我们这里直接使用并查集来实现即可。

2、那么此时我们分类讨论：

①如果N个人，每个人都不会任何一种官方语言，那么对应我们将这N个人每个人都教会同一种语言即可，输出N。

②如果非上述情况，那么对应使得两个集合能够合并在一起，那么使得两个集合中任意一个集合中的任意一个人学会另外一个集合中任意一个人会的语言即可。

那么此时输出集合数-1.

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int f[150];
int map[150][150];
int find(int a)
{
    int r=a;
    while(r!=f[r])
    {
        r=f[r];
    }
    int i=a;
    int j;
    while(i!=r)
    {
        j=f[i];
        f[i]=r;
        i=j;
    }
    return r;
}
void merge(int x,int y)
{
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
    {
        f[xx]=yy;
    }
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(map,0,sizeof(map));
        for(int i=0;i<n;i++)f[i]=i;
        int flag=0;
        for(int  i=0;i<n;i++)
        {
            int k;
            scanf("%d",&k);
            if(k>0)flag=1;
            while(k--)
            {
                int x;
                scanf("%d",&x);
                x--;
                map[i][x]=1;
            }
        }
        if(flag==0)
        {
            printf("%d\n",n);
            continue;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                for(int k=0;k<m;k++)
                {
                    if(map[i][k]==map[j][k]&&map[i][k]==1)
                    {
                        merge(i,j);
                    }
                }
            }
        }
        int output=0;
        for(int i=0;i<n;i++)
        {
            if(f[i]==i)output++;
        }
        printf("%d\n",output-1);
    }
}