首页 > ACM题库 > HDU-杭电 > HDU 1150 Machine Schedule-分治-[解题报告] cpp:nogutter:collapse
2013
12-03

HDU 1150 Machine Schedule-分治-[解题报告] cpp:nogutter:collapse

Machine Schedule

问题描述 :

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.

输入:

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.

输出:

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

样例输入:

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

样例输出:

3

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1150

题目大意:机器调度问题,有两台机器,分别有n和m种工作状态,现有k个任务,需要用这两种机器来完成,由于每更换一次机器的工作状态都需要重启,因此,想要怎样安排这些任务,使得机器的重启次数最少

思路:由于每个任务都可以用A或B机器的某种工作状态来完成,因此,对于每个任务,可以把A,B的工作状态作为顶点,在对应的A和B的工作状态间连线,则对于每条边的两个顶点,因为每个任务都要完成,故至少要有一个顶点(某个机器的一种状态)存在,这样这个问题就转化成了在这些顶点中选取尽量少的顶点(尽量少的工作状态,尽量少的重启次数),使得每一条边都至少和其中一个点相关联,即最小点覆盖问题。由于二分图的最小点覆盖数=最大匹配数,故可用匈牙利算法解决。

#include<stdio.h>   
#include<string.h>   
#define N 110   
int map[N][N],used[N],match[N],n,m;       
int find(int k)     
{   
    int i;   
    for(i=1;i<m;i++) 
    {   
        if(map[k][i] && !used[i])  
        {   
            used[i]=1; 
            if(match[i]==0 || find(match[i]))   
            {   
                match[i]=k;   
                return 1;    
            }   
        }   
    }   
    return 0;   
}   
int main()   
{   
    int i,k,count,x,y;      
    while(scanf("%d",&n),n!=0)
	{
		scanf("%d%d",&m,&k);
		memset(map,0,sizeof(map));  /*第一次忘了这句初始化,wrong了*/
		while(k--)
		{
			scanf("%d%d%d",&i,&x,&y);
			if(x && y)
				map[x][y]=1;
		}
        count=0;  /*记录匹配数*/  
        memset(match,0,sizeof(match));  /*初始时匹配边为空*/     
        for(i=1;i<n;i++)   
        {   
            memset(used,0,sizeof(used));   /*所有顶点置未访问标记*/  
            if(find(i))   
                count++;  /*若找到i的匹配边,则计数器加一*/     
        }   
        printf("%d/n",count);   
    }   
    return 0;   
}