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 committeeYour 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 NCount1 Student1 1 Student1 2 ... Student1 Count1Count2 Student2 1 Student2 2 ... Student2 Count2...... CountP StudentP 1 StudentP 2 ... StudentP CountPThe 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

23 33 1 2 32 1 21 13 32 1 32 1 31 1

Sample Output

YESNO 

Code:

 1 #include <bits/stdc++.h> 2 using namespace std; 3  4 #define CLR(x , v)      memset(x , v , sizeof(x)) 5 static const int MAXN = 1005; 6  7 bool course[MAXN][MAXN]; 8 bool vis[MAXN]; 9 int link[MAXN];10 int p , n;11 int t;12 13 bool Find(int x)14 {15     for(int i = 1 ; i <= n ; ++i)16     {17         if(!vis[i] && course[x][i])18         {19             vis[i] = 1;20             if(!link[i] || Find(link[i]))21             {22                 link[i] = x;23                 return 1;24             }25         }26     }27     return 0;28 }29 30 int main()31 {32     scanf("%d" , &t);33     while(t--)34     {35         CLR(course , 0);36         CLR(vis , 0);37         CLR(link , 0);38 39         scanf("%d%d" , &p , &n);40         for(int i = 1 ; i <= p ; ++i)41         {42             int m;43             scanf("%d" , &m);44             while(m--)45             {46                 int x;47                 scanf("%d" , &x);48                 course[i][x] = 1;49             }50         }51 52         int sum = 0;53 54         for(int i = 1 ; i <= p ; ++i)55         {56             CLR(vis , 0);57             if(Find(i))58                 ++sum;59         }60 61         if(sum == p)62             puts("YES");63         else64             puts("NO");65     }66 }

View Code

HDOJ1083-Courses(二分图匹配(匈牙利算法))