poj题目1469 COURSES （二分图匹配，匈牙利算法）

喜欢ヅ旅行 2022-12-19 06:15 145阅读 0赞

COURSES

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 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;15655</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;6185</td> 
  </tr> 
 </tbody> 
</table>

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

一共有N个学生跟P门课程,一个学生可以任意选一门或多门课,问是否达成:  
1.每个学生代表的都是不同的课（如果一个学生选修的那门课，那么他就可以代表这门课）  
2.每门课都有一个代表

输入为:  
P N(课程数跟学生数)  
接着有P行,格式为Count studenti studenti+1 ……studentcount(Count表示对课程1感兴趣的学生数,接着有Count个学生)  
如第一行2 1 2表示学生1跟学生2对课程1感兴趣  
输出为:  
若每门课都能找到一位代表则输出”YES”,否则为”NO”

假如有三个学生跟三门课程,学生1,2,3.为了跟学生区分,假设3个课程为4,5,6  
左边节点是学生,右边节点是课程,下图表示,学生1对课程4,5感兴趣,学生2对课程5,6感兴趣,学生3对课程6感兴趣

于是问题就变为在二分图中寻找最大匹配,只要这个最大匹配大于或等于课程数P,那么就达到要求了.

#include<stdio.h>
    #include<string.h>
    int m,n;
    int G[110][310],link[310];
    bool vis[310];
    
    int dfs(int x){
    	int i;
    	for(i=1;i<=m;i++){
    		if(G[x][i] && !vis[i]){
    			vis[i]=1;
    			if(link[i]==-1 || dfs(link[i])){
    				link[i]=x;
    				return 1;
    			}
    		}
    	}
    	return 0;
    }
    int main()
    {
    	int t,i,j,sum,num,tmp;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d%d",&n,&m); 
    		sum = 0;
    		memset(G,0,sizeof(G));
    		memset(link,-1,sizeof(link));
    		for(i=1;i<=n;i++){
    			scanf("%d",&num);
    			for(j=1;j<=num;j++){
    				scanf("%d",&tmp);
    				G[i][tmp]=1;
    			}
    		}
    		for(i=1;i<=n;i++){
    			memset(vis,0,sizeof(vis));
    			if(dfs(i)==1) sum++;
    		}
    		printf("%s\n",sum==n?"YES":"NO");
    	}
    	return 0;
    }