（最小点覆盖 匈牙利算法）--pojMachine Schedule

Machine Schedule
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 15751 Accepted: 6755
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

Beijing 2002

思路：
源于http://blog.csdn.net/c20180630/article/details/70209929

先看匈牙利算法：
http://blog.csdn.net/kaisa158/article/details/48718403

我们要将此题转化为匈牙利算法的模型，如下图

例如样例中的输入 0 1 1 1 1 2
就等价于 A中的1号节点与 B中的1号节点相连，A中的1号节点与B中的2号节点相连，…..
到这里之后就相当于要把所有的点给覆盖掉，选择一条边的任意一个点都可以将该边覆盖，用匈牙利算法就可以解决

/*    最小点覆盖：用最少的点覆盖所有的边 */#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int mp[1005][1005];int vis[1005];int link[1005];int n,m,k;int Find(int x){    for(int i=1; i<=m; i++)    {        //有联系  并且未被查找过         if(mp[x][i] && !vis[i])        {            vis[i]=1;            //i没有被匹配，或者  能腾出匹配位置的             if(!link[i] || Find(link[i]) )            {                link[i]=x;                return 1;            }        }    }     return 0;}int main(void){    while(scanf("%d",&n)&&n)    {        cin>>m>>k;        int ans=0;        memset(mp,0,sizeof mp);        memset(link,0,sizeof link);        while(k--)        {            int a,b,c;            cin>>a>>b>>c;            mp[b][c]=1;        }        //匈牙利算法         for(int i=1;i<=n;i++)        {            memset(vis,0,sizeof(vis));            if(Find(i))                ans++;        }        printf("%d\n",ans);    }}