hdu 1150 Machine Schedule

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4793    Accepted Submission(s): 2368

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)

 

 

Recommend

Ignatius.L   |   We have carefully selected several similar problems for you:  1507 1528 1054 1498 1083 

 

 感想：

  开始想了很久，没想到二分图的思想...后来找资料看了，发现自己理解错题意了，题意大概是这样的(网摘)：

  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。

思路(网摘，懒得写)：


二分图的最小顶点覆盖数=最大匹配数

本题就是求最小顶点覆盖数的。



每个任务建立一条边。

最小点覆盖就是求最少的点可以连接到所有的边。本题就是最小点覆盖=最大二分匹配数。

注意一点就是：题目说初始状态为0，所以如果一个任务有一点为0的边不要添加。因为不需要代价


代码：

//最小顶点覆盖==最大匹配数 
//0MS    1200K    909B
#include<stdio.h>
#include<string.h>
const int N=1005;
bool g[N][N],vis[N];
int match[N];
int n,m,k;
bool dfs(int cur)
{
    for(int i=0;i<m;i++){
        if(g[cur][i] && !vis[i]){
            vis[i]=true;
            if(match[i]==-1 || dfs(match[i])){
                match[i]=cur;
                return true;
            }
        }
    }
    return false; 
}
int main(void)
{
    int ik,x,y;
    while(scanf("%d",&n),n)
    {
        scanf("%d%d",&m,&k);
        memset(g,false,sizeof(g));
        memset(match,-1,sizeof(match));
        for(int i=0;i<k;i++){
            scanf("%d%d%d",&ik,&x,&y);
            if(x && y) //这里要注意，A、B一开始都是mode_0 
            g[x][y]=true;
        }
        int ans=0;
        for(int i=0;i<n;i++){
            memset(vis,false,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}