codeforces 277 A Learning Languages
来源:互联网 发布:苹果电脑剪辑软件 编辑:程序博客网 时间:2024/06/04 19:10
A. 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).
Examples
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.
【分析】
裸并查集。
注意t全部为0的情况,答案就是n
【代码】
//codeforces 277 A(AC)#include<iostream>#include<cstdio>#include<vector>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;bool b[101];int father[101],t[101],ans;vector <int> v[101];int find(int x){ if(x==father[x]) return x; return father[x]=find(father[x]);}int main(){ int i,j,k,n,m,x; scanf("%d%d",&n,&m); fo(i,1,100) father[i]=i; fo(i,1,n) { scanf("%d",&t[i]); fo(j,1,t[i]) { scanf("%d",&x); v[x].push_back(i); } } fo(i,1,n) if(t[i]) break; if(i>n) { printf("%d\n",n); return 0; } fo(i,1,m) //枚举语言 { x=v[i].size()-2; fo(j,0,x) { int r1=find(v[i][j]); int r2=find(v[i][j+1]); if(r1!=r2) father[r1]=r2; } } fo(i,1,n) { if(!b[find(i)]) ans++; b[find(i)]=1; } printf("%d\n",ans-1); return 0;}
- [Codeforces] 277A - Learning Languages
- codeforces 277 A Learning Languages
- 【Codeforces Round #170】Codeforces 277A Learning Languages
- CodeForces 277A Learning Languages 并查集
- CodeForces 277A Learning Languages (并查集)
- CodeForces 277A Learning Languages (并查集)
- CodeForces 170 A. Learning Languages //搜索
- CodeForces-Learning Languages
- A. Learning Languages
- 【Codeforces Round #170 Div. 1】 227A Learning Languages
- Codeforces 278C. Learning Languages
- codeforce round 170 A. Learning Languages
- Codeforces 170C Learning Languages (并查集求连通分支)
- Codeforces 278C. Learning Languages 图的遍历
- Codeforces 278C Learning Languages(并查集)
- codeforces 278C. Learning Languages(并查集)
- Codeforces 278C Learning Languages【并查集】水题
- C. Learning Languages
- Docker网络详解及pipework源码解读与实践
- Android中布局总结
- 掌握 Linux 调试技术
- cordova插件
- 常用工具
- codeforces 277 A Learning Languages
- Neo4j使用shell脚本执行命令
- Codeforces Round #361 (Div. 2)——B. Mike and Shortcuts(BFS+小坑)
- Spring Filter过滤器,Spring拦截未登录用户权限限制(转)
- Mybatis缓存粗显理解
- performSelector的简单用法
- 【TCP/IP协议】java SOCKET网络编程
- 艺术品交易平台需求设计文档
- CamShift 目标跟踪算法研究(转)