Consider a group of N students and P courses. Each
student visits zero, one or more than one courses. Your task is to determine
whether it is possible to form a committee of exactly P students that satisfies
simultaneously the conditions:
- every student in the committee represents a different course (a student
can represent a course if he/she visits that course)
- each course has a representative in the committee
Your program should read sets of data from the std
input. The first line of the input contains the number of the data sets. Each
data set is presented in the following format:
P N
Count1
Student1 1 Student1 2 ... Student1
Count1
Count2 Student2 1 Student2
2 ... Student2 Count2
...
CountP
StudentP 1 StudentP 2 ... StudentP
CountP
The first line in each data set contains two positive
integers separated by one blank: P (1 <= P <= 100) - the number of courses
and N (1 <= N <= 300) - the number of students. The next P lines describe
in sequence of the courses ?from course 1 to course P, each line describing a
course. The description of course i is a line that starts with an integer Count
i (0 <= Count i <= N) representing the number of students visiting course
i. Next, after a blank, you抣l find the Count i students, visiting the course,
each two consecutive separated by one blank. Students are numbered with the
positive integers from 1 to N.
There are no blank lines between
consecutive sets of data. Input data are correct.
The result of the program is on the standard
output. For each input data set the program prints on a single line "YES" if it
is possible to form a committee and "NO" otherwise. There should not be any
leading blanks at the start of the line.
1 //272K 422MS C++ 1097B 2014-06-03 12:39:56
2 #include<iostream>
3 #include<vector>
4 #define N 305
5 using namespace std;
6 vector<int>V[N];
7 int vis[N];
8 int match[N];
9 int n;
10 int dfs(int u)
11 {
12 for(int i=0;i<V[u].size();i++){
13 int v=V[u][i];
14 if(!vis[v]){
15 vis[v]=1;
16 if(match[v]==-1 || dfs(match[v])){
17 match[v]=u;
18 return 1;
19 }
20 }
21 }
22 return 0;
23 }
24 int hungary()
25 {
26 memset(match,-1,sizeof(match));
27 int ret=0;
28 for(int i=1;i<=n;i++){
29 memset(vis,0,sizeof(vis));
30 ret+=dfs(i);
31 }
32 return ret;
33 }
34 int main(void)
35 {
36 int t,m,k,a;
37 scanf("%d",&t);
38 while(t--)
39 {
40 scanf("%d%d",&n,&m);
41 for(int i=0;i<=n;i++) V[i].clear();
42 for(int i=1;i<=n;i++){
43 scanf("%d",&k);
44 while(k--){
45 scanf("%d",&a);
46 V[i].push_back(a);
47 //V[b].push_back(a);
48 }
49 }
50 if(hungary()==n) puts("YES");
51 else puts("NO");
52 }
53 return 0;
54 }
