Codeforces 170C Learning Languages (并查集求连通分支)
来源:互联网 发布:java b2c商城源码 编辑:程序博客网 时间:2024/06/04 17:58
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).
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.
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).
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
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.
用并查集来求连通分支,也就是没有共同语言的团体数,当然这里需要注意特判,当所有人都没有共同语言的时候。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=101;const int maxm=101;int cnt[maxm];int peo[maxm][maxn];int pre[maxn],n,m;void init(){ int i; for(i=1; i<=n; i++) pre[i] = i;}int find(int x){ int r=x; while(r!=pre[r]) r=pre[r]; int i=x,j; while(i!=r) { j=pre[i]; pre[i]=r; i=j; } return r;}void unon(int a,int b){ int at = find(a); int bt = find(b); if(at==bt) return; else { pre[at] = pre[bt]; }}int main(){ int k,v; while(~scanf("%d%d",&n,&m)) { memset(peo,0,sizeof(peo)); memset(cnt,0,sizeof(cnt)); init(); bool f=false; for(int i=1; i<=n; i++) { scanf("%d",&k); if(k) f=true; while(k--) { scanf("%d",&v); peo[v][cnt[v]++]=i; } } for(int i=1; i<=m; i++) for(int j=1,u=peo[i][0]; j<cnt[i]; j++) unon(peo[i][j],u); int res=-1; for(int i=1; i<=n; i++) if(pre[i]==i) res++; if(f) printf("%d\n",res); else printf("%d\n",res+1); } return 0;}
- Codeforces 170C Learning Languages (并查集求连通分支)
- Codeforces 278C Learning Languages(并查集)
- codeforces 278C. Learning Languages(并查集)
- Codeforces 278C Learning Languages【并查集】水题
- Codeforces 278C. Learning Languages
- Codeforces Round #170 (Div. 2)---C. Learning Languages(并查集)
- codeforce C. Learning Languages(并查集)
- CodeForces 170 A. Learning Languages //搜索
- CodeForces-Learning Languages
- CodeForces 277A Learning Languages 并查集
- CodeForces 277A Learning Languages (并查集)
- CodeForces 277A Learning Languages (并查集)
- Codeforces 278C. Learning Languages 图的遍历
- C. Learning Languages
- 【Codeforces Round #170】Codeforces 277A Learning Languages
- [Codeforces] 277A - Learning Languages
- codeforces 277 A Learning Languages
- 【Codeforces Round #170 Div. 1】 227A Learning Languages
- CSMA/CD
- github删除repository
- smb
- date命令
- C#继承
- Codeforces 170C Learning Languages (并查集求连通分支)
- DRAM和SRAM特征
- 输出信息到日志(更新版)
- Oracle多实例配置
- cal 命令
- excel使用
- POJ 1410 判断线段相交点在多边形内外
- mysql修改root用户密码
- grep 命令