poj 1469 （hdu1083）COURSES 最大匹配

COURSES

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

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：一共有N个学生跟P门课程,一个学生可以任意选一

门或多门课,问是否达成:

1.每个学生选的都是不同的课(即不能有两个学生选同一门课)

2.每门课都有一个代表(即P门课都被成功选过)

注意：学生可有没选上课的。

匈牙利算法求最大匹配：

#include"stdio.h"
#include"string.h"
#define N 305
int g[N][N],link[N];
int mark[N],n,p;
int find(int k)
{
    int i;
    for(i=1;i<=n;i++)
    {
        if(g[k][i]&&!mark[i])
        {
            mark[i]=1;
            if(!link[i]||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,T,m,v;
    scanf("%d",&T);
    while(T--)
    {
        memset(g,0,sizeof(g));
        memset(link,0,sizeof(link));
        scanf("%d%d",&p,&n);
        for(i=1;i<=p;i++)
        {
            scanf("%d",&m);
            while(m--)
            {
                scanf("%d",&v);
                g[i][v]=1;
            }
        }
        int ans=0;
        for(i=1;i<=p;i++)
        {
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        if(ans==p)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}