HDU 1083  Courses （二分匹配）

Courses

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768K (Java/Others)
Total Submission(s): 1655 Accepted Submission(s):743

Problem Description

Consider a group of N students and Pcourses. Each student visits zero, one or more than one courses.Your task is to determine whether it is possible to form acommittee of exactly P students that satisfies simultaneously theconditions:

. every student in the committee represents a different course (astudent 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 firstline of the input file contains the number of the data sets. Eachdata 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 integersseparated by one blank: P (1 <= P <=100) - the number of courses and N (1 <= N<= 300) - the number of students. The next P linesdescribe in sequence of the courses . from course 1 to course P,each line describing a course. The description of course i is aline that starts with an integer Count i (0 <= Counti <= N) representing the number of students visitingcourse 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. Inputdata are correct.

The result of the program is on the standard output. For each inputdata set the program prints on a single line "YES" if it ispossible to form a committee and "NO" otherwise. There should notbe any leading blanks at the start of the line.

An example of program input and output:

Sample Input

2 3 3 3 1 23 2 1 2 1 1 3 3 2 1 3 2 1 3 1 1

Sample Output

YESNO

Source

Southeastern Europe 2000

 

也算是一个比较典型的二分匹配吧，拿来练练刚学的算法

 

代码：

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

intvis[301],flag[301],map[301][301],n,p;
int DFS(int x)
{
    inti,j;
   for(i=1;i<=map[x][0];i++)
    {
       j=map[x][i];
       if(!vis[j])
       {
           vis[j]=1;
           if(flag[j]==-1 || DFS(flag[j]))
           {
               flag[j]=x;
               return 1;
           }
       }
    }
    return0;
}
int main()
{
    inti,j,t,num;
   scanf("%d",&t);
   while(t--)
    {
       scanf("%d%d",&p,&n);
       for(i=0;i<p;i++)
       {
           scanf("%d",&num);
           map[i][0]=num;
           for(j=1;j<=num;j++)
               scanf("%d",&map[i][j]);
       }
       memset(flag,-1,sizeof(flag));
       for(i=0;i<p;i++)
       {
           memset(vis,0,sizeof(vis));
           if(!DFS(i))
               break;
       }
       if(i<p)
           printf("NO\n");
       else printf("YES\n");
    }
    return0;
}