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.

InputThe 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. 
OutputThe 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

大意:两个机器,分别有N个和M个模式,K个工作,每个工作可以用第一个机器的某个模式完成或者用第二个机器的某个模式完成。每次机器切换模式都需要重新启动。
问最少重启几次机器。
这是一个裸的最小点覆盖,还是求一个图中与每条边相邻的最小点集。最大匹配=最小点覆盖
 #include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
int read(){
int xx=,ff=;char ch=getchar();
while(ch>''||ch<''){if(ch=='-')ff=-;ch=getchar();}
while(ch>=''&&ch<=''){xx=(xx<<)+(xx<<)+ch-'';ch=getchar();}
return xx*ff;
}
const int maxn=;
int lin[maxn],len,N,M,K,ans;
struct edge{
int y,next;
}e[];
inline void insert(int xx,int yy){
e[++len].next=lin[xx];
e[len].y=yy;
lin[xx]=len;
}
int tim,pretim,match[maxn],vis[maxn];
bool hun(int x){
for(int i=lin[x];i;i=e[i].next)
if(vis[e[i].y]<=pretim){
vis[e[i].y]=++tim;
if(match[e[i].y]==||hun(match[e[i].y])){
match[e[i].y]=x;
match[x]=e[i].y;
return ;
}
}
return ;
}
int main(){
//freopen("in","r",stdin);
//freopen("out","w",stdout);
while(){
N=read();
if(!N)
break;
M=read(),K=read();
memset(lin,,sizeof(lin));len=;
for(int i=;i<=K;i++){
read();
int t1=read(),t2=read();
insert(t1,t2+N);
}
ans=;tim=;pretim=;
memset(match,,sizeof(match));
memset(vis,,sizeof(vis));
for(int i=;i<=N+M;i++)
if(!match[i]){
pretim=tim;
vis[i]=++tim;
if(hun(i))
ans++;
}
printf("%d\n",ans);
}
return ;
}
 

hdu1150——最小点覆盖的更多相关文章

  1. HDU1150 Machine Schedule(二分图最大匹配、最小点覆盖)

    As we all know, machine scheduling is a very classical problem in computer science and has been stud ...

  2. 【hdu1150】【Machine Schedule】二分图最小点覆盖+简单感性证明

    (上不了p站我要死了,侵权度娘背锅) 题目大意 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ai,如 ...

  3. ACM/ICPC 之 机器调度-匈牙利算法解最小点覆盖集(DFS)(POJ1325)

    //匈牙利算法-DFS //求最小点覆盖集 == 求最大匹配 //Time:0Ms Memory:208K #include<iostream> #include<cstring&g ...

  4. 【POJ 3041】Asteroids (最小点覆盖)

    每次选择清除一行或者一列上的小行星.最少选择几次. 将行和列抽象成点,第i行为节点i+n,第j列为节点j,每个行星则是一条边,连接了所在的行列. 于是问题转化成最小点覆盖.二分图的最小点覆盖==最大匹 ...

  5. POJ 2226 最小点覆盖(经典建图)

    Muddy Fields Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8881   Accepted: 3300 Desc ...

  6. nyoj 237 游戏高手的烦恼 二分匹配--最小点覆盖

    题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=237 二分匹配--最小点覆盖模板题 Tips:用邻接矩阵超时,用数组模拟邻接表WA,暂时只 ...

  7. [USACO2005][POJ2226]Muddy Fields(二分图最小点覆盖)

    题目:http://poj.org/problem?id=2226 题意:给你一个字符矩阵,每个位置只能有"*"或者“.",连续的横着或者竖的“*"可以用一块木 ...

  8. POJ3041Asteroids(最小点覆盖+有点小抽象)

    Asteroids Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18289   Accepted: 9968 Descri ...

  9. hdu 1054 最小点覆盖

    Sample Input 4 0:(1) 1 1:(2) 2 3 2:(0) 3:(0) 5 3:(3) 1 4 2 1:(1) 0 2:(0) 0:(0) 4:(0)   Sample Output ...

随机推荐

  1. 探索java世界中的日志奥秘

                    java日志简单介绍 对于一个应用程序来说日志记录是必不可少的一部分.线上问题追踪,基于日志的业务逻辑统计分析等都离不日志.JAVA领域存在多种日志框架,目前常用的日志 ...

  2. C#——反射动态创建类的实例

    “反射”其实就是利用程序集的元数据信息. 反射可以有很多方法,编写程序时请先导入 System.Reflection 命名空间. 若要反射当前项目中的类(即当前项目已经引用它了),可以使用下面的写法. ...

  3. 使用 Spring Social 连接社交网络

    Spring Social 框架是spring 提供社交平台的分享组件 https://www.ibm.com/developerworks/cn/java/j-lo-spring-social/

  4. 学习java编程能往哪些方向发展

    当下Java训练非常热,是因为通过学习java能够快速的就业,这对于今年就业压力非常大的大学生来说,无疑是一条就业的捷路,虽然培教育费动辄过万,但还是非常值得的. 可是你可曾想过,学习了java编程后 ...

  5. linux 的sed命令解释 sed ':t;N;s/\n/,/;b t' 将换行符换成逗号

    linux 的sed命令解释 sed ':t;N;s/\n/,/;b t' 将换行符换成逗号 实现的功能是吧换行符换成逗号了,自己试验过. 求解释,:t N b t 都是什么意思??? :t 定义la ...

  6. 使用Python的Flask框架,结合Highchart,动态渲染图表

    服务端动态渲染图表 参考文章链接:https://www.highcharts.com.cn/docs/dynamic-produce-html-page 参考文章是使用php写的,我这边改用pyth ...

  7. PAT 1106 Lowest Price in Supply Chain

    A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyone invo ...

  8. ACdream 1032 Component

    Component Time Limit: 5000ms Memory Limit: 64000KB This problem will be judged on ACdream. Original ...

  9. 【ACM】hdu_zs3_1007_Rails_201308100802

    Rails Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)Total Submissi ...

  10. 关于Spring的xml文档的简单实用配置

    Spring的spring.xml文档的配置 最近在写Spring的配置文件时,发现Spring文档的配置其实没必要那么繁琐记忆,网上的很多文章都写得很繁琐,如果所有的东西按照路径去查找,可以很快的帮 ...