hdu 1150 Machine Schedule（求最小点覆盖）

Machine Schedule

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)

二分图中有König定理，定理如下

最大分配=最小点覆盖

证明过程http://www.matrix67.com/blog/archives/116

此题我们可以将同一个任务的AB点相连，这样题目就转化成了求最小点覆盖，因为只要一条边被覆盖了说明，一个任务在AB中的其中一个点加入了最小点覆盖中，即任务已完成

#include<cstdio>
#include<cstring>
using namespace std;
#define N 150
int map[N][N],used[N],girl[N];//used记录在每一轮搜索中该点有无被访问过，girl记录与右边点配对的左边点
int n,m;
int find(int i)
{
    int j;
    for(j=0;j<m;j++)
    {
        if(map[i][j]&&!used[j])
        {
            used[j]=1;
            if(girl[j]==0||find(girl[j]))//如果girl没匹配过或者girl的男人能另寻一个girl则更新
            {
                girl[j]=i;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int k,i,cnt,a,b;
    while(scanf("%d",&n)&&n)
    {
        memset(girl,0,sizeof(girl));
        memset(map,0,sizeof(map));
        scanf("%d%d",&m,&k);
        while(k--)
        {
            scanf("%d%d%d",&i,&a,&b);
            if(a&&b) map[a][b]=1;//初始是0的话就没必要计算了
        }
        cnt=0;
        for(i=0;i<n;i++)
        {
            memset(used,0,sizeof(used));
            if(find(i)) cnt++;//如果成功更新，说明匹配对数增加
        }
        printf("%d\n",cnt);
    }
    return 0;
}