版权声明:来自: 码代码的猿猿的AC之路 http://blog.csdn.net/ck_boss https://blog.csdn.net/u012797220/article/details/32732003

最小点覆盖=最大匹配

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5350    Accepted Submission(s): 2650

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
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; const int maxn=1200; int n,m,p,linker[maxn];
bool mp[maxn][maxn],used[maxn]; bool dfs(int u)
{
for(int i=0;i<m;i++)
{
if(mp[u][i]&&!used[i])
{
used[i]=true;
if(linker[i]==-1||dfs(linker[i]))
{
linker[i]=u;
return true;
}
}
}
return false;
} int hungary()
{
int ret=0;
memset(linker,-1,sizeof(linker));
for(int i=0;i<n;i++)
{
memset(used,false,sizeof(used));
if(dfs(i)) ret++;
}
return ret;
} int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
scanf("%d%d",&m,&p);
memset(mp,false,sizeof(mp));
for(int i=0;i<p;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(b&&c) mp[b][c]=true;
}
printf("%d\n",hungary());
}
return 0;
}

HDOJ 1150 Machine Schedule的更多相关文章

  1. hdoj 1150 Machine Schedule【匈牙利算法+最小顶点覆盖】

    Machine Schedule Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  2. hdu 1150 Machine Schedule(最小顶点覆盖)

    pid=1150">Machine Schedule Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/327 ...

  3. 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/ ...

  4. 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 ...

  5. 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? ...

  6. 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? ...

  7. hdu 1150 Machine Schedule (二分匹配)

    Machine Schedule Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  8. HDU——1150 Machine Schedule

    Machine Schedule Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  9. 二分图最大匹配(匈牙利算法)简介& Example hdu 1150 Machine Schedule

    二分图匹配(匈牙利算法) 1.一个二分图中的最大匹配数等于这个图中的最小点覆盖数 König定理是一个二分图中很重要的定理,它的意思是,一个二分图中的最大匹配数等于这个图中的最小点覆盖数.如果你还不知 ...

随机推荐

  1. es6的...用法

    ...将一个数组转为用符号分隔的参数序列 1.console.log(1, ...[2, 3, 4], 5) // 1 2 3 4 5 2. var args = [0, 1, 2]; f.apply ...

  2. php curl文件上传

    <?php /** * 这是一个自动化部署的类, 非常简单,思想就是压缩,上传,然后解压覆盖,所以请小心使用. * @author liuchao <249757247@qq.com> ...

  3. Alexa TOP 100万的域名列表

    Alexa是一家专门发布网站世界排名的网站,是亚马逊公司的一家子公司.Alexa每天在网上搜集多达几十亿的网址链接,而且为其中的每一个网站进行了排名. Alexa通过Alexa官网查询好像TOP 50 ...

  4. SQL的一对多,多对一,一对一,多对多什么意思?

    1.一对多:比如说一个班级有很多学生,可是这个班级只有一个班主任.在这个班级中随便找一个人,就会知道他们的班主任是谁:知道了这个班主任就会知道有哪几个学生.这里班主任和学生的关系就是一对多. 2.多对 ...

  5. release,debug库互调用,32位,64位程序与库互调用

    以下是基于visual studio 2015和cmake的实验 1,debug或release的应用程序都可以调用release的库2,win32和x64的应用和库无法互调用,在VS中链接时会有一堆 ...

  6. yield(放弃、谦逊、礼让) - 瞬时的,暂时放了马上再抢

    两个线程抢占CPU各自执行任务,代码如下: public class Demo03 { public static void main(String[] args) throws Interrupte ...

  7. 【转】python---方法解析顺序MRO(Method Resolution Order)<以及解决类中super方法>

    [转]python---方法解析顺序MRO(Method Resolution Order)<以及解决类中super方法> MRO了解: 对于支持继承的编程语言来说,其方法(属性)可能定义 ...

  8. PHP模拟请求和操作响应

    模拟请求 fsockopen <?php // 建立连接 $link = fsockopen('localhost', '80'); define('CRLF', "\r\n" ...

  9. mysql多对多查询 原生写法

    准备工作,1.创建表 CREATE TABLE IF NOT EXISTS `users` ( `id` INTEGER NOT NULL AUTO_INCREMENT, `name` VARCHAR ...

  10. python 递归,深度优先搜索与广度优先搜索算法模拟实现

    一.递归原理小案例分析 (1)# 概述 递归:即一个函数调用了自身,即实现了递归 凡是循环能做到的事,递归一般都能做到! (2)# 写递归的过程 1.写出临界条件 2.找出这一次和上一次关系 3.假设 ...