POJ 1469：COURSES

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19458		Accepted: 7658

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

选课。给了每一门课有那些学生选。要问的是在给定的关系的情况下，能不能求出一个学生集合，这个集合里面能够满足这样的关系：

1.每门课都能派出一个学生代表。

2.每个学生都代表了一门课。

不能更匈牙利了。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int grid[305][305];
int link[305];
int visit[305];
int n,k,V1,V2;
int result;

bool dfs(int x)
{
	int i;
	for(i=1;i<=V2;i++)
	{
		if(grid[x][i]==1&&visit[i]==0)
		{
			visit[i]=1;
			if(link[i]==-1||dfs(link[i]))
			{
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}

bool Magyarors()
{
	int i;

	result=0;
	memset(link,-1,sizeof(link));//！！这里不能是0

	for(i=1;i<=V1;i++)
	{
		memset(visit,0,sizeof(visit));
		if(!dfs(i))
		{
			return false;
		}
	}
	return true;
}

int main()
{
	int Test,i,j,num,temp;
	cin>>Test;

	while(Test--)
	{
		memset(grid,0,sizeof(grid));
		cin>>V1>>V2;

		for(i=1;i<=V1;i++)
		{
			scanf("%d",&num);
			for(j=1;j<=num;j++)
			{
				scanf("%d",&temp);
				grid[i][temp]=1;
			}
		}
		if(V1>V2)
		{
			cout<<"NO"<<endl;
		}
		else
		{
			if(Magyarors())
			{
				cout<<"YES"<<endl;
			}
			else
			{
				cout<<"NO"<<endl;
			}
		}
	}
	return 0;
}

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