hdu-----（1150）Machine Schedule（最小覆盖点）

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5817    Accepted Submission(s): 2932

Problem Description

As
we all know, machine scheduling is a very classical problem in computer
science and has been studied for a very long history. Scheduling
problems differ widely in the nature of the constraints that must be
satisfied and the type of schedule desired. Here we consider a 2-machine
scheduling problem.

There are two machines A and B. Machine A
has n kinds of working modes, which is called mode_0, mode_1, …,
mode_n-1, likewise machine B has m kinds of working modes, mode_0,
mode_1, … , mode_m-1. At the beginning they are both work at mode_0.

For
k jobs given, each of them can be processed in either one of the two
machines in particular mode. For example, job 0 can either be processed
in machine A at mode_3 or in machine B at mode_4, job 1 can either be
processed in machine A at mode_2 or in machine B at mode_4, and so on.
Thus, for job i, the constraint can be represent as a triple (i, x, y),
which means it can be processed either in machine A at mode_x, or in
machine B at mode_y.

Obviously, to accomplish all the jobs, we
need to change the machine's working mode from time to time, but
unfortunately, the machine's working mode can only be changed by
restarting it manually. By changing the sequence of the jobs and
assigning each job to a suitable machine, please write a program to
minimize the times of restarting machines.

 

Input

The
input file for this program consists of several configurations. The
first line of one configuration contains three positive integers: n, m
(n, m < 100) and k (k < 1000). The following k lines give the
constrains of the k jobs, each line is a triple: i, x, y.

The input will be terminated by a line containing a single zero.

 

Output

The output should be one integer per line, which means the minimal times of restarting machine.

 

Sample Input

5 5 10
0 1 1
1 1 2
2 1 3
3 1 4
4 2 1
5 2 2
6 2 3
7 2 4
8 3 3
9 4 3
0

 

Sample Output

3

 

Source

Asia 2002, Beijing (Mainland China)

 

代码：  最小点覆盖=最大匹配....

 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 using namespace std;
 int const maxn=;
 int n,m,k;
 bool mat[maxn][maxn];
 bool vis[maxn];
 int mac[maxn];
 bool match(int x)
 {
     for(int i=;i<=m;i++)
     {
         if(mat[x][i]&&!vis[i]){
             vis[i]=;
             if(!mac[i]||match(mac[i])){
                 mac[i]=x;
                 return ;
             }
         }
     }
     return ;
 }
 int main(){
     int a,b,c;
     //freopen("test.in","r",stdin);
     while(scanf("%d",&n),n!=){
          memset(mat,,sizeof(mat));
          memset(mac,,sizeof(mac));
        scanf("%d%d",&m,&k);
        for(int i=;i<k;i++){
          scanf("%d%d%d",&a,&b,&c);
          mat[b][c]=;
        }
        int ans=;
      for(int i=;i<=n;i++){
          memset(vis,,sizeof(vis));
          if(match(i))ans++;
      }
      printf("%d\n",ans);
     }
   return ;
 }