Codeforces 278C. Learning Languages
来源:互联网 发布:linux私房菜 第四版 编辑:程序博客网 时间:2024/06/05 02:47
Description
The "BerCorp" company has got n employees. These employees can usem approved official languages for the formal correspondence. The languages are numbered with integers from1 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 costs1 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 andm (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 thei-th line is integer ki (0 ≤ ki ≤ m) — the number of languages thei-th employee knows. Next, the i-th line contains ki integers —aij (1 ≤ aij ≤ m) — the identifiers of languages thei-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 Input
5 51 22 2 32 3 42 4 51 5
0
8 703 1 2 31 12 5 42 6 71 32 7 41 1
2
2 21 20
1
Hint
In the second sample the employee 1 can learn language2, and employee 8 can learn language4.
In the third sample employee 2 must learn language2.
题意:公司有N个员工,每个人可能会M种语言,要想让所有人能直接或间接沟通,至少需要学多少人次。
解法:只需要用并差集找出已经形成集合的个数n,那么还需要n-1就能把整个合并成一个集合,然后加上一门
语言都不会的人数即可,需要注意磁体有特数据,所有人会的语言数都为0的情况。
#include<iostream>#include<cstdio>#include<cstring>#include<set>using namespace std;const int maxn=101;int n,m,fa[maxn];set<int> s[maxn];int find(int x){ return x==fa[x] ? x : fa[x]=find(fa[x]);}bool check(int i,int j){ set<int>::iterator x=s[i].begin(); while(x!=s[i].end()){ int y=*x; if(s[j].find(y)!=s[j].end()) return true; x++; //cout << 1 << endl; } return false;}int main(){ scanf("%d %d",&n,&m); for(int i=0;i<=n;i++) fa[i]=i; int sum,x,ans=0; for(int i=0;i<n;i++){ scanf("%d",&sum); if(sum==0) ans++; while(sum--){ scanf("%d",&x); s[i].insert(x); } } if(ans==n) printf("%d\n",n); else{ for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) if(check(i,j)){ int fi=find(i),fj=find(j); if(fi!=fj) fa[fi]=fj; } int ant=0; for(int i=0;i<n;i++) if(fa[i]==i) ant++; printf("%d\n",ant-1); } return 0;}
- Codeforces 278C. Learning Languages
- 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
- CodeForces-Learning Languages
- Codeforces 170C Learning Languages (并查集求连通分支)
- C. Learning Languages
- [Codeforces] 277A - Learning Languages
- codeforces 277 A Learning Languages
- CodeForces 170 A. Learning Languages //搜索
- Codeforces Round #170 (Div. 2)---C. Learning Languages(并查集)
- codeforce C. Learning Languages(并查集)
- 【Codeforces Round #170】Codeforces 277A Learning Languages
- CodeForces 277A Learning Languages 并查集
- CodeForces 277A Learning Languages (并查集)
- 【Codeforces Round #170 Div. 1】 227A Learning Languages
- CodeForces 277A Learning Languages (并查集)
- 负载均衡(LB)
- 平衡二叉树
- 灵活组装Json的数据使用Gson的JsonParser和JsonReader解析Json详解例子
- Transforming Cumulative Ceilometer Stats to Gauges
- 查看linux系统环境
- Codeforces 278C. Learning Languages
- Hdu 1527 取石子游戏 (威佐夫博弈)
- 实验项目2-7:素因子分解
- hd 1789 Doing Homework again
- Android--选择对话框--AlertDialog
- spring jdbc:initialize-database使用详解
- 阅读Logback文档笔记--Logback的Encoder配置
- 复杂度nlog(n)之堆排序
- eclipse下SpringBoot开发和测试