hdoj 1083 Courses【匈牙利算法】

Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4669    Accepted Submission(s): 2230

Problem 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

Your program should read sets of data from a text file. The first line of the input file contains the number of the data sets. Each data set is presented in the following format:

P N
Count1 Student1 1 Student1 2 ... Student1 Count1
Count2 Student2 1 Student2 2 ... Student2 Count2
......
CountP StudentP 1 StudentP 2 ... StudentP 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'll 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.

An example of program input and output:

 

 

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

看到英文题就头疼，看不懂啊  伤心..........

  此题是说如果每门课都要有人选且选此课的人不能选择其他课则输出yes否则输出no

用匈牙利算法  如果最后配对出来的总对数等于总的课数则正确

#include<stdio.h>
#include<string.h>
#define MAX 1100
int cour,stu,p;
int map[MAX][MAX];
int vis[MAX],s[MAX];
int find(int x)
{
	int i,j;
	for(i=1;i<=stu;i++)
	{
		if(map[x][i]&&vis[i]==0)//如果学生对这门课程感兴趣且 没被标记 
		{          //（这里被标记就是说第i个学生选上了这门课） 
			vis[i]=1;
			if(s[i]==0||find(s[i]))//如果第i个学生没有选上课或者可以换课 
			{
				s[i]=x;//则让第i个学生选上这门课 
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int i,j,k,t,sum;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&cour,&stu);
		memset(map,0,sizeof(map));
		memset(s,0,sizeof(s));
		for(i=1;i<=cour;i++)
		{
			scanf("%d",&p);
			while(p--)
			{
				scanf("%d",&k);
				map[i][k]=1;//给对应课程和对应学生标记 
			}
		}
		sum=0;
		for(i=1;i<=cour;i++)
		{
			memset(vis,0,sizeof(vis));
			if(find(i))
			sum++;
		}
		if(sum==cour)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}