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.
InputThe 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.
OutputThe 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 题意：有两台机器A和B以及N个需要运行的任务。每台机器有M种不同的模式，而每个任务都恰好在一台机器上运行。如果它在机器A上运行，则机器A需要设置为模式xi,如果它在机器B上运行，则机器A需要设置为模式yi。每台机器上的任务可以按照任意顺序执行，但是每台机器每转换一次模式需要重启一次。请合理为每个任务安排一台机器并合理安排顺序，使得机器重启次数尽量少（kuangbin大佬的权威解释，比较通俗易懂，就借用一下了，望大佬莫怪罪啊）。 思路：每个任务的两个模式之间建一条边，很显然构成了一张二分图，现在我们需要使重启次数最小，相当于求该图的最小点覆盖。
根据二分图的特性--最小点覆盖=最大匹配数，可以轻松做出此题。不过因为初始状态为零，所以如果有模式为0的边就不添加了） 代码如下：
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std; int match[][],vis[],n,m;
int ye[]; bool find(int x)
{
for(int i=;i<=m;i++)
{
if(match[x][i]==&&!vis[i]) //如果存在x到i的边且该边没有被匹配
{
vis[i]=; //匹配这条边，将该边设为已被访问
if(ye[i]==-) //如果该边没有匹配边，设置x为该边的匹配边
{
ye[i]=x;
return true;
}
else
{
if(find(ye[i])==true)//如果该边被匹配了，寻找边ye[i]是否能与其他边匹配，如果能，空出i的位置
{
ye[i]=x;
return true;
}
}
}
}
return false;
} int main()
{
int k,i,x,y;
int ans;
while(~scanf("%d",&n))
{
ans=;
memset(match,,sizeof(match));
memset(ye,-,sizeof(ye));
if(n==)
{
break;
}
scanf("%d%d",&m,&k);
while(k--)
{
scanf("%d%d%d",&i,&x,&y);
match[x][y]=;
}
for(int j=;j<=n;j++)
{
memset(vis,,sizeof(vis));
if(find(j)==true)
{
ans++;
}
}
printf("%d\n",ans);
}
return ;
}
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