POJ 1469 COURSES

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22584		Accepted: 8914

Description

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

Input

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 Student_{1 1} Student_{1 2} ... Student_{1 Count1}
Count2 Student_{2 1} Student_{2 2} ... Student_{2 Count2}
...
CountP Student_{P 1} Student_{P 2} ... Student_{P 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.

Output

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.

Sample Input

Sample Output

YES
NO

Source

Southeastern Europe 2000

题目分析：该题给出P门课程，N个学生，问能否从中选出P个学生，使每个学生上不同的课，且每个课程有一个学生。典型的二分图匹配的问题，我们只要计算最大二分图匹配数，如果和课程数相同就输出YES，否则输出NO。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 bool g[110][310];
 6 bool flag,visit[310];
 7 int match[310];
 8 int p,n;
 9 bool dfs(int u){
10     for(int i=1;i<=n;++i){
11         if(g[u][i]&&!visit[i]){
12             visit[i]=true;
13             if(match[i]==-1 || dfs(match[i])){
14                 match[i]=u;
15                 return true;
16             }
17         }
18     }
19     return false;
20 }
21 int main() {
22     int i,j,k,t,v,ans;
23     scanf("%d",&t);
24     while(t--){
25         scanf("%d %d", &p, &n);
26         for(i=1;i<=p;i++){
27             for(j=1;j<=n;j++)
28                 g[i][j]=false;
29         }
30         for(i=1;i<=n;i++)
31             match[i]=-1;
32         flag=true;
33         for(i=1;i<=p;i++){
34             scanf("%d",&k);
35             if(k == 0) flag=false;
36             while(k--){
37                 scanf("%d",&v);
38                 g[i][v]=true;
39             }
40         }
41         if(flag){
42             ans=0;
43             for(i=1;i<=p;i++){
44                 memset(visit,false,sizeof(visit));   //清空上次搜索时的标记
45                 if(dfs(i)) ans++;
46             }
47             if(ans == p)puts("YES");
48             else puts("NO");
49         }
50         else puts("NO");
51     }
52     return 0;
53 }