HDU——1150 Machine Schedule

Posted 2020-10-04

　　　　　　　　　　　　Machine Schedule

　　　　　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　Total Submission(s): 9461    Accepted Submission(s): 4755

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

大致题意：有两台机器A和B以及k个需要运行的任务。每台机器有n,m种不同的模式，而每个任务都恰好能在一台机器上运行。 机器A上有模式 mode_0, mode_1, …, mode_n-1,机器B上有模式： mode_0, mode_1, … , mode_m-1。 开始的工作模式都是mode_0。每个任务有对应的运行模式，(i, x, y)表示i任务对应的A B机器上的运行模式mode_x, mode_y。 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。并求其值。

思路：

最小点覆盖（最大匹配数、、）

我们对于每一个任务连边，然后我们现在要完成任务并且我们要要求机器重启的次数最少，那么也就是说我们需要用到最少的点把所有的边连起来，这样就转化成了最小点覆盖的裸题了

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 1010
using namespace std;
bool vis[N];
int n,m,k,x,y,z,ans,girl[N],map[N][N];
int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
int find(int x)
{
    for(int i=1;i<=m;i++)
    {
        if(!vis[i]&&map[x][i])
        {
            vis[i]=true;
            if(girl[i]==-1||find(girl[i])) {girl[i]=x; return 1;}
        }
    }
    return 0;
}
int main()
{
    while(1)
    {
        n=read();if(n==0) break;
        m=read(),k=read();ans=0;
        memset(map,0,sizeof(map));
        memset(girl,-1,sizeof(girl));
        for(int i=1;i<=k;i++)
        {
            z=read(),x=read(),y=read();
            map[x][y]=1;
        }
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(find(i)) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}