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 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：给出课程数跟学生数，以及课程对应的学生。
求是否存在让所有课程跟不同的学生对应。


#include <stdio.h>
#include <stdlib.h>
#define maxn 310
int map[maxn][maxn], hash[maxn], pre[maxn], gn, gm;
//gn左边x的点数，gm右边y的点数 
//pre是右边节点连接x中的节点 
//hash是右边的节点有没有被访问过 
void set()
{
    for (int i = 1; i <= gn; i++)
    {
        for (int j = 1; j <= gm; j++)
        {
            map[i][j] = 0;
        }
    }
    for (int j = 1; j <= gm; j++)
    {
        pre[j] = -1;
    }
}
int han(int u)
{
    for (int v = 1; v <= gm; v++)
    {
        if (hash[v] == 0 && map[u][v])//边存在且右节点没被访问过 
        {
            hash[v] = 1;
            if (pre[v] == -1 || han(pre[v]))//v还没有前置节点或v的前置节点存在增广路 
            {
                pre[v] = u;//设定v的前置节点 
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int t, i, j, m, u, v, num;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &gn, &gm);
        set();
        for (i = 1; i <= gn; i++)
        {
            scanf("%d", &m);
            while (m--)
            {
                scanf("%d", &v);
                map[i][v] = 1;
            }
        }
        for (i = 1, num = 0; i <= gn; i++)
        {
            for (j = 1; j <= gm; j++)
            {
                hash[j] = 0;
            }
            if (han(i))
            {
                num++;
            }
        }
        if (num == gn)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    system("pause");
    return 0;
}