8.15 F

F - COURSES

 Consider a group of N students and Pcourses. Each student visits zero, one or more than one courses. Your task isto determine whether it is possible to form a committee of exactly P studentsthat satisfies simultaneously the conditions: 

 

·        every student in the committee represents a differentcourse (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 setis 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 oneblank: 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 thecourses �from course 1 tocourse P, each line describing a course. The description of course i is a linethat starts with an integer Count i (0 <= Count i <= N) representing thenumber of students visiting course i. Next, after a blank, you抣l find the Count i students, visiting thecourse, each two consecutive separated by one blank. Students are numbered withthe positive integers from 1 to N. 
There are no blank lines between consecutive sets of data. Input data arecorrect. 

Output

The result of the program is on the standard output. Foreach input data set the program prints on a single line "YES" if itis possible to form a committee and "NO" otherwise. There should notbe any leading blanks at the start of the line.

Sample Input

2

3 3

3 1 2 3

2 1 2

1 1

3 3

2 1 3

2 1 3

1 1

Sample Output

YES

NO

 题意：有p门课程，n个学生，每个学生在一个committee代表一门课程，求学生和课程的最大匹配，若最大匹配等于p，输出YES，否则输出NO。

思路：先将数据读成一张图，用匈牙利算法计算计算连通情况，最后判断所有的点是否连通，是的话输出YES，否则输出NO。

#include<stdio.h>#include<string.h>int a[600][600],macth[600],book[600];int m,n;int dfs(int x)//匈牙利算法{int i;for(i=1;i<=m;i++){if(book[i]==0&&a[x][i]==1){book[i]=1;if(macth[i]==-1||dfs(macth[i])){macth[i]=x;return 1;}}}return 0;}int main(){int i,j,k,t1,t2,sum,t,s,s1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(a,0,sizeof(a));memset(macth,-1,sizeof(macth));for(i=1;i<=n;i++){scanf("%d",&s);for(j=0;j<s;j++){scanf("%d",&s1);a[i][s1]=1;}}/*for(i=1;i<=n;i++){for(j=1;j<=m;j++){printf("%d ",a[i][j]);}printf("\n");}*/sum=0;for(i=1;i<=n;i++){memset(book,0,sizeof(book));if(dfs(i))sum++;//记录有多少点连通}if(sum==n)//判断是否所有的科目都有课代表printf("YES\n");elseprintf("NO\n");}return 0;}