A. Learning Languages
time limit per test:2 seconds
memory limit per test:256 megabytes


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.

题目链接:http://codeforces.com/problemset/problem/277/A

题目大意:n个员工。m种语言。每一个员工会k种语言,问总共还要学习多少语言,能够让每两个员工能够直接或者间接交流

题目分析:非常明显的并查集问题,注意当k等于0时,该员工必学一门语言。具体见程序凝视

#include <cstdio>
#include <cstring>
int a[105], fa[105]; void UF_set()
{
for(int i = 0; i < 105; i++)
fa[i] = i;
} int Find(int x)
{
return x == fa[x] ? x : fa[x] = Find(fa[x]);
} void Union(int a, int b)
{
int r1 = Find(a);
int r2 = Find(b);
if(r1 != r2)
fa[r2] = r1;
} int main()
{
int n, m, cnt = 0, ans = 0; //ans表示理论上须要学习的人数,cnt表示没人学的
scanf("%d %d", &n, &m);
memset(a, 0, sizeof(a));
UF_set();
for(int i = 0; i < n; i++)
{
int k, fir, next;
scanf("%d", &k);
if(k == 0) //若一门语言都不会。则必要学一门
{
ans++;
continue;
}
scanf("%d", &fir);
a[fir] ++;
for(int i = 1; i < k; i++)
{
scanf("%d", &next);
a[next]++;
Union(fir, next);
}
}
for(int i = 1; i <= m; i++)
{
if(a[i] == 0) //记录没人学的
cnt++;
if(fa[i] == i) //记录集合个数
ans++;
}
//没人学的必定自成一个集合,我们要把它减去,由于既然没一个人学
//它就不须要再被学,若ans等于m说明每两个人之间不能交流。则每人都要学习
//一门语言,否则拿集合个数减去没人学的语言个数再减1,这里减1的含义是
//随意n个集合,n-1条线就能将其连通
printf("%d\n", (cnt == m) ? n : ans - cnt - 1);
}

版权声明:本文博主原创文章。博客,未经同意不得转载。

CodeForces 277A Learning Languages (并检查集合)的更多相关文章

  1. [Codeforces Round #170 Div. 1] 277A Learning Languages

    A. Learning Languages time limit per test:2 seconds memory limit per test:256 megabytes input standa ...

  2. Codeforces 278C Learning Languages(并查集)

    题意抽象出来就是求联通块的个数吧,然后添加最少边使图联通. 注意所有人都不会任何语言的时候,答案是n而不是n-1. #include<algorithm> #include<iost ...

  3. BZOJ3296: [USACO2011 Open] Learning Languages

    3296: [USACO2011 Open] Learning Languages Time Limit: 5 Sec  Memory Limit: 128 MBSubmit: 81  Solved: ...

  4. HDU 1272 小希迷宫(并检查集合)

    意甲冠军:被判处无向图无环和连接无处不在 思考:并检查集合,trap 您可能有一个直接输入0 0 并且....合并的时候按某一个方向会爆栈,爆了好几次...下次考虑一下直接递归找祖先吧 #includ ...

  5. BZOJ3296:Learning Languages(简单并查集)

    3296: [USACO2011 Open] Learning Languages Time Limit: 5 Sec  Memory Limit: 128 MBSubmit: 436  Solved ...

  6. C. Learning Languages 求联通块的个数

    C. Learning Languages 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring&g ...

  7. codeforces 277 A Learning Languages 【DFS 】

    n个人,每个人会一些语言,两个人只要有会一门相同的语言就可以交流,问为了让这n个人都交流,至少还得学多少门语言 先根据n个人之间他们会的语言,建边 再dfs找出有多少个联通块ans,再加ans-1条边 ...

  8. hdu1325 Is It A Tree?并检查集合

    pid=1325">职务地址 试想一下,在词和话题hdu1272是一样的. 可是hdu1272的博文中我也说了.数据比較水,所以我用非并查集的方法就AC了. 可是这题的数据没那么水,要 ...

  9. URAL - 1966 - Cycling Roads(并检查集合 + 判刑线相交)

    意甲冠军:n 积分,m 边缘(1 ≤ m < n ≤ 200),问:是否所有的点连接(两个边相交.该 4 点连接). 主题链接:http://acm.timus.ru/problem.aspx? ...

随机推荐

  1. 怎么确定你的CPU是否支持64位虚拟化

    http://www.grc.com/securable.htm 第一个64位表示你的电脑最多支持多少位的系统,32或者64. 第二个表示你的硬件是否支持DEP?Yes,支持.No,不支持.OFF,表 ...

  2. template method pattern

    //DataViewer.cs using System; namespace TemplateMethodSample { abstract class DataViewer { //抽象方法:获取 ...

  3. [转载] 树莓派读取温湿度传感器DHT11

    原文地址: http://blog.csdn.net/liang890319/article/details/8739683 硬件: 树莓派 2.0 DHT模块  接树莓派5V GND GPIO1 功 ...

  4. 几种流行Webservice控制框架

     转会[http://blog.csdn.net/thunder4393/article/details/5787121],写的非常好,以收藏. 1      摘要 开发webservice应用程序中 ...

  5. POJ 1236 Network of Schools(强连通分量)

    POJ 1236 Network of Schools 题目链接 题意:题意本质上就是,给定一个有向图,问两个问题 1.从哪几个顶点出发,能走全全部点 2.最少连几条边,使得图强连通 思路: #inc ...

  6. $.ajax通路RESTful Web Service一个错误:Unsupported Media Type

    最近项目,使用头版jquery ajax访问背景CXF发布时间rest维修,结果遇到了错误"Unsupported Media Type". 公布的服务java代码例如以下: im ...

  7. CentOS在安装配置 Ngnix_tomcat_PHP_Mysql

    安装Nginx yum install nginx 假设显示找不到 nginx包,新建一个文件/etc/yum.repos.d/nginx.repo,内容: [nginx] name=nginx re ...

  8. PHP PDO sqlite ,Unable to Open database file的解决方法

    t.php在网站的根目录. fdy.db在inc文件夹下; t.php中sqlite路径写成相对路径 $db = new PDO('sqlite:inc/fdy.db'); 开始提示 Fatal er ...

  9. 探索Windows Azure 监控和自动伸缩系列2 - 获取虚拟机的监控定义和监控数据

    上一篇博文介绍了如何连接Windows Azure: http://www.cnblogs.com/teld/p/5113063.html 本篇我们继续上次的示例代码,获取虚拟机的监控定义和监控数据. ...

  10. HDU 4513 哥几个系列故事——形成完善II manacher求最长回文

    标题来源:哥几个系列故事--形成完善II 意甲冠军:中国 思维:在manacher断 保证非严格递减即可了 #include <cstdio> #include <cstring&g ...