CodeForces 277A Learning Languages 并查集

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博客介绍了如何利用并查集解决在给定员工和语言数量的情况下，找到公司最小花费使所有员工能相互沟通的问题。通过分析每个人会的语言，将相同语言的员工归为一类，并计算需要学习语言的员工数量。

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
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
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

有n个人,m种语言.给出每个人会的语言(也可能一种都不会),问最少让几个人学语言,可以使得大家可以互相沟通.

解题思路

利用并查集,发现会同一种语言的就把他们放到一起.最后发现par[i] = i的情况有两种.一种是一种语言都不会的,还有就是会语言的.发现如果会语言的集合<=1时,有多少不会语言的就需要有几个学语言的.直接输出即可.
如果会语言的集合>1时,需要再加上会语言的集合再-1.
(换句话说,可以用不会语言的集合+会语言的集合-1,但是全都是不会语言的集合的时候需要特判)

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 110;
int mem[maxn];//mem[i]表示i语言上次出现在哪个人
int par[maxn];
int f[maxn];
int n,m;
int _find(int x)
{
    if(x == par[x]) return x;
    return par[x] = _find(par[x]);
}
void _unite(int a,int b)
{
    a = _find(a);
    b = _find(b);
    if(a != b) par[a] = b;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1 ; i <= n ; i ++) par[i] = i;
    for(int i = 1 ; i <= n ; i ++) {
        int q;
        scanf("%d",&q);
        if(q == 0) f[i] = 1;
        while(q--) {
            int a;
            scanf("%d",&a);
            if(mem[a] != 0) {_unite(mem[a],i);continue;}
            mem[a] = i;
        }
    }
    int cnt = 0;
    for(int i = 1; i <= n ; i ++) if(par[i] == i && f[i]) cnt++;
    int _cnt = 0;
    for(int i = 1 ; i <= n ; i ++) if(par[i] == i && !f[i]) _cnt++;
    if(_cnt > 1) cnt += (_cnt-1);
    printf("%d\n",cnt);
    return 0;
}