1469 COURSES //MAXMATCH

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COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7838		Accepted: 3113

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

#include<stdio.h>
#include<string.h>
int usedif[105];
//usedif[i]记录Y顶点子集中编号为i的顶点是否使用,注意Y子集中的最多顶点数为(7-1)*12+12=84
int link[105];//link[i]记录与Y顶点子集中编号为i的顶点相连的X顶点子集中x的编号
int mat[305][105];//mat[i][j]表示顶点i与j之间是否有边
int gx,gy;//gx为X顶点子集中的顶点数目,gy为Y顶点子集中的顶点数目
bool can(int t) //判断X中的顶点t在Y中是否有顶点与之匹配
{
    for(int i=1; i<=gy; i++) //注意范围
    {
        if(usedif[i]==0&&mat[t][i]) //Y中的顶点i未匹配且t与i之间有边
        {
            usedif[i]=1;
            if(link[i]==-1||can(link[i])) //link[i]=-1表示顶点i还未匹配
//can(link[i])表示顶点i已经匹配而现在唯一的路径,就是走到与i节点匹配的顶点link[i]处继续进行匹配
            {
                link[i]=t;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch()
{
    int num=0;
    memset(link,-1,sizeof(link));
    for(int i=1; i<=gx; i++) //对X中的每个顶点在Y中寻找与其匹配的顶点，注意范围
    {
        memset(usedif,0,sizeof(usedif));
        if(can(i)) num++;
    }
    return num;//返回最大匹配数
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int p,n,t;
        while(scanf("%d%d",&p,&n)!=EOF)
        {
            gx=n;
            gy=p;
            memset(mat,0,sizeof(mat));
            for(int i=1; i<=p; i++)
            {
                scanf("%d",&t);
                for(int j=0; j<t; j++)
                {
                    int x;
                    scanf("%d",&x);
                    mat[x][i]=1;
                }
            }
            if(n>=p&&MaxMatch()==p) printf("YES/n");
            else printf("NO/n");
        }
    }
    return 0;
}