HDU-----（1083）Courses（最大匹配）

Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3767    Accepted Submission(s): 1800

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

 

Source

Southeastern Europe 2000

  给定一个课程数目和每一个门课程听课的人的编号，问能否每一门课都有人听课...!

代码：

 #include<cstdio>
 #include<cstring>
 #include<vector>
 using namespace std;
 const int maxn=;
 vector<int>mat[maxn];
 int judge[maxn];
 bool vis[maxn];
 bool skm(int x){
   vector<int>::iterator it;
   for(it=mat[x].begin();it<mat[x].end();it++)
     {
       if(!vis[*it])
       {
         vis[*it]=;
      if(judge[*it]==||skm(judge[*it]))
      {
          judge[*it]=x;
          return ;
      }
     }
   }
   return ;
 }

 int main()
 {
     int cas,n,m,t,a,i;
     scanf("%d",&cas);
     while(cas--)
     {
      scanf("%d%d",&n,&m);
      memset(judge,,sizeof(int)*(m+));
     for( i=;i<=n;i++)
     {
       mat[i].clear();
       scanf("%d",&t);
      while(t--){
          scanf("%d",&a);
         mat[i].push_back(a);
      }
     }
     for(i=;i<=n;i++){
       memset(vis,,sizeof(bool)*(m+));
       if(!skm(i))break;
     }
     if(i>n) printf("YES\n");
     else printf("NO\n");
     }
  return ;
 }

采用邻接矩阵做：

 #include<cstdio>
 #include<cstring>
 #include<vector>
 #include<cstdlib>
 #include<iostream>
 using namespace std;
 const int maxn=;
 bool mat[maxn][maxn];
 int judge[maxn];
 bool vis[maxn];
 int n,m;
 bool skm(int x){
   for(int i=;i<=m;i++)
   {
       if(!vis[i]&&mat[x][i])
       {
         vis[i]=;
      if(judge[i]==||skm(judge[i]))
      {
          judge[i]=x;
          return ;
      }
     }
   }
   return ;
 }
 int main()
 {
     int cas,t,a,i;
     //freopen("test.in","r",stdin);
     scanf("%d",&cas);
     while(cas--)
     {
      scanf("%d%d",&n,&m);
      memset(judge,,sizeof(judge));
      memset(mat,,sizeof(mat));
     for( i=;i<=n;i++)
     {
       scanf("%d",&t);
      while(t--)
      {
          scanf("%d",&a);
          mat[i][a]=;
      }
     }
     for(i=;i<=n;i++){
       memset(vis,,sizeof(vis));
       if(!skm(i))break;
     }
     if(i>n)
         printf("YES\n");
     else
         printf("NO\n");
     }
  return ;
 }