• hdu 1150 Machine Schedule(二分匹配,简单匈牙利算法)


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

    Machine Schedule

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6733    Accepted Submission(s): 3375


    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
     
    题目大意:有两台机器A和B,A机器有n种工作方式,B机器有m种工作方式。共有k个任务。每个任务恰好在一条机器上运行。
    如果任务在A机器上运行,就需要转换为模式Xi,如果在B机器上运行,就需要转换为模式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。
     
    解题思路:
    把机器A的N种模式作为二分图的左部,机器B的M种模式作为二分图的右部,如果某个任务可以使用机器A的模式xi也可以使用机器B的模式yi完成,则连接xi,yi。
    题目要求使机器重启的次数要尽量少,又要把所有的任务都执行完,也就可以把题目转换成最小顶点覆盖,根据二分图的性质:最小顶点覆盖=最大匹配数。

     

    详见代码。
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 int Map[110][110],vis[110],n,m;
     8 int ok[110];
     9 
    10 bool Find(int x)
    11 {
    12     for (int i=1;i<=m;i++)
    13     {
    14         if (Map[x][i]==1&&!vis[i])
    15         {
    16             vis[i]=1;
    17             if (ok[i]==-1)
    18             {
    19                 ok[i]=x;
    20                 return true;
    21             }
    22             else
    23             {
    24                 if (Find(ok[i])==true)
    25                 {
    26                     ok[i]=x;
    27                     return true;
    28                 }
    29             }
    30         }
    31     }
    32     return false;
    33 }
    34 
    35 int main()
    36 {
    37     int k,i,x,y;
    38     int ans;
    39     while (~scanf("%d",&n))
    40     {
    41         ans=0;
    42         memset(Map,0,sizeof(Map));
    43         memset(ok,-1,sizeof(ok));
    44         if (n==0)
    45             break;
    46         scanf("%d%d",&m,&k);
    47         while (k--)
    48         {
    49             scanf("%d%d%d",&i,&x,&y);
    50             //if(x>0&&y>0)
    51                 Map[x][y]=1;
    52         }
    53         for (int j=1;j<=n;j++)
    54         {
    55             memset(vis,0,sizeof(vis));
    56             if (Find(j)==true)
    57                 ans++;
    58         }
    59         printf ("%d
    ",ans);
    60     }
    61     return 0;
    62 }
  • 相关阅读:
    libcurl进行HTTP GET获取JSON数据(转载)
    c/c++ 动态申请数组(转载)
    利用注册表写开机启动程序(转载)
    VC++使用IMAPI调用Outlook邮箱客户端和Foxmail邮箱客户端遇到的问题
    解决Duilib集成CEF浏览器在Win10无法向客户区拖拽文件
    解决往监控目录拖拽文件夹无法监控到的问题
    VC++ 实现修改文件创建、访问、修改时间属性(转载)
    Windows中的时间(SYSTEMTIME和FILETIME) (转载)
    ThinkPHP真正疑难问题笔记
    git版本控制管理实践-4
  • 原文地址:https://www.cnblogs.com/qq-star/p/4675600.html
Copyright © 2020-2023  润新知