题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1150
Machine Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6733 Accepted Submission(s):
3375
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.
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.
the minimal times of restarting machine.
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int Map[][],vis[],n,m;
int ok[]; bool Find(int x)
{
for (int i=;i<=m;i++)
{
if (Map[x][i]==&&!vis[i])
{
vis[i]=;
if (ok[i]==-)
{
ok[i]=x;
return true;
}
else
{
if (Find(ok[i])==true)
{
ok[i]=x;
return true;
}
}
}
}
return false;
} int main()
{
int k,i,x,y;
int ans;
while (~scanf("%d",&n))
{
ans=;
memset(Map,,sizeof(Map));
memset(ok,-,sizeof(ok));
if (n==)
break;
scanf("%d%d",&m,&k);
while (k--)
{
scanf("%d%d%d",&i,&x,&y);
//if(x>0&&y>0)
Map[x][y]=;
}
for (int j=;j<=n;j++)
{
memset(vis,,sizeof(vis));
if (Find(j)==true)
ans++;
}
printf ("%d\n",ans);
}
return ;
}