hdu 1150 Machine Schedule(二分匹配,简单匈牙利算法)
来源:互联网 发布:js验证提示信息 编辑:程序博客网 时间:2024/06/05 02:50
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.
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.
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)
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 else23 {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\n",ans);60 }61 return 0;62 }
0 0
- hdu 1150 Machine Schedule(二分匹配,简单匈牙利算法)
- HDU 1150:Machine Schedule(二分匹配,匈牙利算法)
- HDU 1150 Machine Schedule(二分匹配+匈牙利算法)
- hdu 1150 Machine Schedule(二分匹配,匈牙利算法)
- Machine Schedule(二分图匹配--匈牙利算法)
- HDU 1150 Machine Schedule(匈牙利算法 二分图的最小顶点覆盖 二分图最大匹配)
- POJ 1325-Machine Schedule(二分图匹配-匈牙利算法)
- hdu 1150 Machine Schedule(匈牙利算法)
- hdu 1150 Machine Schedule ( 二分匹配)
- hdu 1150 Machine Schedule(二分匹配)
- HOJ 1056 Machine Schedule (二分图匹配,匈牙利算法)
- HDU 1150 Machine Schedule匈牙利算法
- HDU 1150 Machine Schedule(匈牙利算法)
- HDU 1150 Machine Schedule (匈牙利算法详解)
- hdu Machine Schedule 1150 二分图匹配
- hdu 1150 Machine Schedule (经典二分匹配)
- HDU 1150 Machine Schedule 二分图匹配
- HDU 1150 Machine Schedule(二分图匹配)
- inner join, left join, right join, full join 的区别?
- Hadoop伪分布配置详解
- 算法之匈牙利算法
- UVA 12545 Bits Equalizer
- hdu 1281 棋盘游戏(二分匹配)
- hdu 1150 Machine Schedule(二分匹配,简单匈牙利算法)
- 随记
- hdu 5326 Work(杭电多校赛第三场)
- su 与su -
- hdu 5319 Painter(杭电多校赛第三场)
- hdu 5328 Problem Killer(杭电多校赛第四场)
- hdu 1151 Air Raid(最小路径覆盖)
- HNU Joke with permutation (深搜dfs)
- hdu 1498 50 years, 50 colors(二分匹配_匈牙利算法)