Problem 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

Your program should read sets of data from a text file. The first line of the input file 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'll 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.

An example of program input and output:
 

 

Sample Input
2
3 3
3 1 2 3
2 1 2
1 1
3 3  
1 3
2 1 3
1 1
 

 

Sample Output
YES
NO
 
 
 

题意和思路：有p门课程，n个学生，每个学生在一个选择代表一门课程，求课程的被选的数量是否等于p，输出YES，否则输出NO。
                  很容易想到什么思路呢？我们可以看成课程来挑选学生，每个课程可以喜欢多个学生，但是只能选择
                  一个学生，问最后是否所有的课程都能找到他自己喜欢的学生。所以我们就用最大匹配来做。貌似直接匈牙利算法就搞定了。
 
 
 
 
 
 
 
 
 
 
 1 #include<iostream> 
 2 #include<algorithm> 
 3 #include<cstdio>  
 4 #include<cstring>  
 5 using namespace std;  
 6 const int maxn = 310;  
 7 int T, p, n, co, st;  
 8 int f[maxn][maxn];  
 9 int s[maxn];  
10 int xh[maxn];  
11 bool flag;  
12 bool PP(int course)//分配学生 
13 {  
14     for(int i =1;i <= n; ++i)  
15     {  
16         if(f[course][i] && !xh[i])  
17         {  
18             xh[i] = 1;  
19             if(!s[i] || PP(s[i]))  
20             {  
21                 s[i] = course;  
22                 return true;  
23             }  
24         }  
25     }  
26     return false;  
27 }  
28 int main()  
29 {  
30     cin >> T;  
31     while(T--)  
32     {  
33         flag = true;  
34         memset(s, 0, sizeof(s));  
35         memset(f, 0, sizeof(f));  
36         scanf("%d %d", &p, &n);  
37         for(int i = 1;i <= p; ++i)  
38         {  
39             scanf("%d", &co);  
40             while(co--)  
41             {  
42                 scanf("%d", &st);  
43                 f[i][st] = 1;  
44             }  
45         }  
46         for(int i = 1;i <= p; ++i)  
47         {  
48             memset(xh, 0, sizeof(xh));  
49             if(!PP(i))  
50             {  
51                 flag = false;  
52                 break;  
53             }  
54         }  
55         if(flag) cout <<"YES\n";  
56         else cout <<"NO\n";  
57     }  
58     return 0;  
59 }