POJ1469:COURSES

最新推荐文章于 2018-01-28 20:49:48 发布

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最新推荐文章于 2018-01-28 20:49:48 发布

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分类专栏： POJ 文章标签：图论二分图

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探讨了如何通过二分图最大匹配算法解决课程与学生间的委员会组建问题，确保每门课程都有代表且代表各不相同。

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点击打开题目链接

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15621		Accepted: 6171

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

=====================================题目大意=====================================

现在有N个学生和P门课程，每个学生掌握了一定门数的课程，求解是否能够组成满足以下两点的委员会：

1、委员会中的每个学生都只能为互不相同的一门课程做代表（只有当掌握了某门课程时学生才能作为该门课程的代表）。

2、每门课程都有委员会中的一个学生作为代表。

=====================================算法分析=====================================

二分图的最大匹配，注意看清题意，这里只是要求所有课程都有匹配的学生，并未要求所有学生都有匹配的课程。

=======================================代码=======================================

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

int T,P,N,Linker[305];

bool Edge[305][305],Vis[305];

bool DFS(int U)
{
	for(int v=1;v<=N;++v)
	{
		if(Edge[U][v]&&!Vis[v])
		{
			Vis[v]=true;
			if(Linker[v]==-1||DFS(Linker[v]))
			{
				Linker[v]=U;
				return true;
			}
		}
	}
	return false;
}

bool Hungary()
{
	memset(Linker,-1,sizeof(Linker));
	for(int u=1;u<=P;++u)
	{
		memset(Vis,0,sizeof(Vis));
		if(!DFS(u)) { return false; }
	}
	return true;
}

int main()
{
	while(scanf("%d",&T)==1) while(T--)
	{
		memset(Edge,0,sizeof(Edge));
		scanf("%d%d",&P,&N);
		for(int i=1;i<=P;++i)
		{
			int count;
			scanf("%d",&count);
			while(count--)
			{
				int student;
				scanf("%d",&student);
				Edge[i][student]=true;
			}
		}
		printf("%s\n",Hungary()?"YES":"NO");
	}
	return 0;
}