POJ 1469 COURSES //简单二分图

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 
  

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. 

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.

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

YES
NO

Southeastern Europe 2000

#include <stdio.h>
#include <string.h>

bool map[105][305];	//图，记录 第i个同学 选修 第j个课程
bool visit[305];	//记录该点是否被访问过
int r_jilu[305];	//记录右边项 对应的 左边点

bool dfs(int n, int r);

int main()
{
	int t;
	int l, r;
	int ans, k, v;
	int i;

	scanf("%d", &t);
	while(t--)
	{
		scanf("%d%d", &l, &r);

		memset(map, false, sizeof(map));
		for(i=1; i<=r; i++)
			r_jilu[i] = -1;		//初始化

		for(i=1; i<=l; i++)
		{
			scanf("%d", &k);
			while(k--)
			{
				scanf("%d", &v);
				map[i][v] = true;
			}
		}

		ans = 0;
		for(i=1; i<=l; i++)
		{
			memset(visit, false, sizeof(visit));
			if( dfs(i, r) )
				ans++;
		}
		if(ans == l)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}
bool dfs(int n, int r)	// 匈牙利算法  
{
	int i;
	for(i=1; i<=r; i++)
	{
		if(map[n][i]==true && visit[i]==false)	//当此点为左边对应的点，并且没被访问过时
		{
			visit[i] = true;
			if(r_jilu[i]==-1 || dfs(r_jilu[i], r))//如果i未在前一个匹配M中，或者i在匹配M中，但是从与i相邻的节点出发可以有增广路径
			{
				r_jilu[i] = n;		//更新
				return true;
			}
		}
	}
	return false;
}