[数据结构与算法]专题训练3 Bproblem—Learning language

题意

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).

思路

这题可以把多种语言看成是一个节点而每一个员工就是把这些节点给连接起来的操作，也就是并查集的解题技巧。而且我们需要判断有哪些语言是没有出现过的，只要是没有出现过的语言就不用计算，剩下就是查找这个多种语言的节点可以分成多少段，然后只要给前一段的某个员工开一门课程关于下一节点的任何语言就可以将两个段给连接在一起。特别情况是出现0种的时候，这个员工是必须开一门课程，来连接到任意一段就行。

代码

#include<bits/stdc++.h> 
using namespace std;
int father[110],vis[110];
int find(int x)
{
 	return x==father[x] ? x:father[x]=find(father[x]);
}
void _union(int x,int y)
{
 	x=find(x);
 	y=find(y);
 	if(x != y)
 	{
  		father[x]=y;
 	}
}
int main()
{
	int n,m;cin>>n>>m;
	for(int i=0;i<=101;i++)
	  	father[i]=i;
	int sum =0 ; 
	for(int i = 0;i < n;i ++)
	{
		int t;cin >> t;
		int a[105];
		if(t==0)
		{
		   	sum ++;
		  	continue;
		}
		cin >> a[0];
		vis[a[0]]++;
		for(int  j = 1;j < t;j ++)
		{
			cin >>a[j];
			vis[a[j]]++;
			_union(a[j-1],a[j]);
		}
	}
	int p = 0;
	for(int i = 1;i <= m;i ++)
	{
		if(vis[i])
		{
			if(father[i]==i)
			{
			   	p++;
			}
		}
	}
	p=max(0,p-1);
	cout<<sum+p<<endl; 
    return 0;
}