codeforce round 170 A. Learning Languages

A. Learning Languages

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

Input

5 5
1 2
2 2 3
2 3 4
2 4 5
1 5

Output

0

Input

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

Output

2

Input

2 2
1 2
0

Output

1

这道题真的很奇怪，想了很久，也知道大概就是并查集。。。可就是想不到具体去，结果还是参考了一下别人的结题报告，才有了思路。。。

最主要的就是要将每个人当成一个节点，将他们懂的语言作为联系将他们连接起来，最后判断有几组并查集。。。

这里有特殊情况，就是有可能每个人懂的语言都不相通，这样的话，他们要学的语言就要比之前的还要多出一种

代码：

#include<iostream>
#include<string.h>
using namespace std;
//int pre[100];
int fa[105];
int a[105][105];
int times[105];
int find(int re){
    if(fa[re]==0){
         return re;
         }
    else
         return fa[re]=find(fa[re]);
}
int totether(int x,int y){
    int pre1=find(x);
    int pre2=find(y);
    if(pre1==pre2)
        return 0;
    else{
        fa[pre1]=pre2;
        return 0;
        }
}
int main(){
    int n,m;
    while(cin>>n>>m){
        int flag=0;
        memset(fa,0,sizeof(fa));
       // memset(pre,0,sizeof(0));
        memset(times,0,sizeof(times));
        for(int i=1;i<=n;i++){
                int temp;
                cin>>temp;
                if(temp>0)
                    flag=1;
                for(int j=0;j<temp;j++){
                        int temp1;
                        cin>>temp1;
                        a[temp1][times[temp1]++]=i;
                        }
                        }
        for(int k=1;k<=m;k++){
                for(int g=1;g<times[k];g++){
                        totether(a[k][g-1],a[k][g]);
                        }
                        }
        int res=0;
        for(int i=1;i<=n;i++){
                if(fa[i]==0){
                      res++;
                      }
                      }
        if(flag==0)
                cout<<res<<endl;
        else
                cout<<res-1<<endl;
                }
     return 0;
     }