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). InputThe 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).
用并查集维护讲同一种语言的人，最后答案就是连通块个数-1。
特判一下没有人会语言的情况，因为这样不能第一次就靠学习别的人的语言来减少连通块个数。
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
vector<int> a[110];
int fa[110],now;
int find(int x)
{return x==fa[x]?x:fa[x]=find(fa[x]);
}
void join(int x,int y)
{int u=find(x),v=find(y);if (u!=v){now--;fa[u]=v;}
}
int main()
{int i,j,k,m,n,x,y,z;bool flag=0;scanf("%d%d",&n,&m);for (i=1;i<=n;i++){scanf("%d",&k);if (k) flag=1;while (k--){scanf("%d",&x);a[x].push_back(i);}}if (!flag){printf("%d\n",n);return 0;}for (i=1;i<=n;i++)fa[i]=i;now=n;for (i=1;i<=m;i++)for (j=1;j<a[i].size();j++)join(a[i][j-1],a[i][j]);printf("%d\n",now-1);
}