hdu 1150 Machine Schedule(最小顶点覆盖)
pid=1150">Machine Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5424 Accepted Submission(s): 2691
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.
x, y.
The input will be terminated by a line containing a single zero.
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
3
题意:有A、B两个机器。每一个机器有多个状态,每一个状态能够完毕某项工作。求完毕一些工作所需的最少状态。
最小顶点覆盖:包括二分图的X,Y部的部分顶点的一个集合,使得全部的边至少有一个顶点在该点集内。
最小顶点覆盖 = 二分图的最大匹配
#include"stdio.h"
#include"string.h"
#define N 105
int g[N][N];
int link[N],mark[N],n,m;
int find(int k)
{
int i;
for(i=1;i<m;i++)
{
if(!mark[i]&&g[k][i])
{
mark[i]=1;
if(!link[i]||find(link[i]))
{
link[i]=k;
return 1;
}
}
}
return 0;
}
int main()
{
int i,a,b,c,k;
while(scanf("%d",&n),n)
{
memset(g,0,sizeof(g));
memset(link,0,sizeof(link));
scanf("%d%d",&m,&k);
while(k--)
{
scanf("%d%d%d",&c,&a,&b);
if(a&&b) //起始状态不需转换
{
g[a][b]=1;
}
}
int ans=0;
for(i=1;i<n;i++)
{
memset(mark,0,sizeof(mark));
ans+=find(i);
}
printf("%d\n",ans);
}
return 0;
}
hdu 1150 Machine Schedule(最小顶点覆盖)的更多相关文章
- 匈牙利算法模板 hdu 1150 Machine Schedule(二分匹配)
二分图:https://blog.csdn.net/c20180630/article/details/70175814 https://blog.csdn.net/flynn_curry/artic ...
- 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/ ...
- hdu - 1150 Machine Schedule (二分图匹配最小点覆盖)
http://acm.hdu.edu.cn/showproblem.php?pid=1150 有两种机器,A机器有n种模式,B机器有m种模式,现在有k个任务需要执行,没切换一个任务机器就需要重启一次, ...
- HDU - 1150 Machine Schedule(二分匹配---最小点覆盖)
题意:有两台机器A和B,A有n种工作模式(0~n-1),B有m种工作模式(0~m-1),两台机器的初始状态都是在工作模式0处.现在有k(0~k-1)个工作,(i,x,y)表示编号为i的工作可以通过机器 ...
- hdu 1150 Machine Schedule 最少点覆盖转化为最大匹配
Machine Schedule Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php? ...
- hdu 1150 Machine Schedule 最少点覆盖
Machine Schedule Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php? ...
- hdu 1150 Machine Schedule (二分匹配)
Machine Schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- hdu 1150 Machine Schedule hdu 1151 Air Raid 匈牙利模版
//两道大水……哦不 两道结论题 结论:二部图的最小覆盖数=二部图的最大匹配数 有向图的最小覆盖数=节点数-二部图的最大匹配数 //hdu 1150 #include<cstdio> #i ...
- HDU——1150 Machine Schedule
Machine Schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
随机推荐
- Adaboost的几个人脸检测网站
[1]基础学习笔记之opencv(1):opencv中facedetect例子浅析 http://www.cnblogs.com/tornadomeet/archive/2012/03/22/2411 ...
- 联系我们_鲲鹏Web数据抓取 - 专业Web数据采集服务提供者
联系我们_鲲鹏Web数据抓取 - 专业Web数据采集服务提供者 首页 > 联系我们 我们的联系方式如下: 029 - 82542052(陕西 西安) 13389148466 或 13571845 ...
- Nginx使用ngx_zeromq模块返回502错误的解决方法
/********************************************************************* * Author : Samson * Date ...
- Jquery利用ajax调用asp.net webservice的各种数据类型(总结篇)
原文:Jquery利用ajax调用asp.net webservice的各种数据类型(总结篇) 老话说的好:好记心不如烂笔头! 本着这原则,我把最近工作中遇到的jquery利用ajax调用web服务的 ...
- LeetCode——Container With Most Water
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). ...
- 修ecshop品牌筛选以LOGO图片形式显示
如何实现商品列表页属性筛选区品牌筛选以LOGO形式展示,最模板总结ecshop/'>ecshop教程入下: 1.修改 category.php 文件,将(大概215行) $sql = " ...
- Cocos2d-x Layout简单使用
1. Text* alert = Text::create("Layout", "fonts/Marker Felt.ttf", 30 ); alert-> ...
- 广东省-IT红黑榜排名公司名称
红榜Top100 Order Company Name Point Change 1 百富计算机技术(深圳)有限公司 94.00 -- 2 中国网通广州分公司 88.00 -- 3 深圳市汇 ...
- 《Effective C++ 》学习笔记——规定10
***************************************转载请注明出处:http://blog.csdn.net/lttree************************** ...
- FS SIP呼叫的消息线程和状态机线程
THREAD 当收到一次呼叫的时候,FS会在TU层创建两个线程,一个线程为状态机线程,另外一个为消息线程.状态机线程通过switch_core_session_thread_launch创建,顾名思义 ...