codeforces 278C. Learning Languages（并查集）

C. 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
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 test(s)

Input

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

Output

0

Input

8 7
03 1 2 3
1 12 5 4
2 6 7
1 3
2 7 4
1 1

Output

2

Input

2 21 20

Output

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.

题目大意是：有n个人，m中语言。每个人各自都多多少少会一种到几种语言（当然也有人一门都不会），然后他们可以直接交流或者间接交流（借助他人）。每个人学一门语言要花费1块钱，问最少要花多少钱，能让他们相互沟通。

那么很容易想到并查集。题目给的数据范围比较小，所以我可以通过搜索前面的人的语言，如果两人都会同一门语言，就把他们放到一个集合里面去。

#include<stdio.h>

#include<string.h>

#define M 1000
int father[M],vis[M];

struct node{

int lang[M];

}p[M];

int findroot(int x)

{

int r=x;

while(r!=father[r])

r=father[r];

int i=x,j;

while(i!=r)

{

j=father[i];

father[i]=r;

i=j;

}

return r;

}

void tomerge(int a,int b)

{

a=findroot(a);b=findroot(b);

if(a!=b)

{

if(a>b)

father[a]=b;

else father[b]=a;

}

}

void tosearch(int x,int i,int k)

{

int j,l=1;

for(j=1;j<i;j++)

{

for(l=1;l<=100;l++)

{if(x==p[j].lang[l]){

tomerge(i,j);return;

}

}

}

}

int main()

{

int n,m,i,j,k[M],a,l,f,t;

while(scanf("%d%d",&n,&m)!=EOF)

{

f=0; t=0;

memset(vis,0,sizeof(vis));

for(i=1;i<=n;i++)

father[i]=i;

for(i=1;i<=n;i++)

{

scanf("%d",&k[i]);

for(j=1;j<=k[i];j++)

{

scanf("%d",&p[i].lang[j]);

if(vis[p[i].lang[j]])tosearch(p[i].lang[j],i,k[i]);

vis[p[i].lang[j]]=1;

}

}

for(i=1;i<=n;i++)

{

// printf("%d ",father[i]);

if(findroot(i)==i)t++;

}

for(i=1;i<=n;i++)

{

if(k[i]!=0)f=1;

}

if(f==1)

printf("%d\n",t-1);

else printf("%d\n",n); //这里有一个特殊情况，如果每个人不会语言，那么每个人都要学。

}

return 0;

}