poj 1469 COURSES（二分匹配模板）

题目链接：http://poj.org/problem?id=1469

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

Source

Southeastern Europe 2000

题意：该题给出P门课程，N个学生，问是否能从中选出P个学生，使每一个学生上不同的课，且每一个课程有一个学生。

典型的二分图匹配的问题。我们仅仅要计算最大二分图匹配数。假设和课程数同样就输出YES，否则输出NO。

代码例如以下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

/* **************************************************************************
//二分图匹配（匈牙利算法的DFS实现）
//初始化：g[][]两边顶点的划分情况
//建立g[i][j]表示i->j的有向边就能够了，是左边向右边的匹配
//g没有边相连则初始化为0
//L是匹配左边的顶点数，R是匹配右边的顶点数
//调用：res=hungary();输出最大匹配数
//长处：适用于稠密图，DFS找增广路，实现简洁易于理解
//时间复杂度:O(VE)
//***************************************************************************/
//顶点编号从0開始的
#define MAXN 317
int p,n;//L,R数目
int k;
int g[MAXN][MAXN], linker[MAXN];
bool used[MAXN];
int dfs(int L)//从左边開始找增广路径
{
	int R;
	for(R = 1 ; R <= n; R++ )//这个顶点编号从0開始，若要从1開始须要改动
	{
		if(g[L][R]!=0 && !used[R])
		{//找增广路，反向
			used[R]=true;
			if(linker[R] == -1 || dfs(linker[R]))
			{
				linker[R]=L;
				return 1;
			}
		}
	}
	return 0;//这个不要忘了。常常忘记这句
}
int hungary()
{
	int res = 0 ;
	int L;
	memset(linker,-1,sizeof(linker));
	for( L = 1; L <= p; L++ )
	{
		memset(used,0,sizeof(used));
		if(dfs(L) != 0)
			res++;
	}
	return res;
}
int main()
{
	int i,j,res,R;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&p,&n);
		memset(g,0,sizeof(g));
		for(i = 1; i <= p; i++)
		{
			scanf("%d",&k);
			for(j = 1 ; j <= k; j++ )
			{
				scanf("%d",&R);
				g[i][R] = 1;
			}
		}
		res = hungary();
		if(res == p)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0 ;
}