ZOJ 1364 Machine Schedule(二分图最大匹配)
题意 机器调度问题 有两个机器A,B A有n种工作模式0...n-1 B有m种工作模式0...m-1 然后又k个任务要做 每一个任务能够用A机器的模式i或b机器的模式j来完毕 机器開始都处于模式0 每次换模式时都要重新启动 问完毕全部任务机器至少重新启动多少次
最基础的二分图最大匹配问题 对于每一个任务把i和j之间连一条边就能够构成一个二分图 那么每一个任务都能够相应一条边 那么如今就是要找最少的点 使这些点能覆盖全部的边 即点覆盖数 又由于二分图的点覆盖数 = 匹配数 那么就是裸的求二分图最大匹配问题了 两边的点数都不超过100直接DFS增广即可了
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int g[N][N], a[N], b[N], vis[N];
int n, m, k, ans; int dfs(int i) //DFS增广
{
for(int j = 1; j < m; ++j)
{
if(g[i][j] && !vis[j])
{
vis[j] = 1;
if( b[j] == -1 || dfs(b[j]))
{
//机器a的模式i与机器b的模式j匹配
a[i] = j;
b[j] = i;
return 1;
}
}
}
return 0;
} int main()
{
int u, v, t;
while(scanf("%d", &n), n)
{
memset(g, 0, sizeof(g));
scanf("%d%d", &m, &k);
for(int i = 0; i < k; ++i)
{
scanf("%d%d%d", &t, &u, &v);
g[u][v] = 1;
} ans = 0;
memset(a, -1, sizeof(a));
memset(b, -1, sizeof(b));
//状态0不须要重新启动 所以能够忽略0
for(int i = 1; i < n; ++i)
{
if(a[i] == -1) //i没被匹配 以i为起点找一条增广路
{
memset(vis, 0, sizeof(vis));
ans += dfs(i); //
}
} printf("%d\n", ans);
}
return 0;
}
//Last modified : 2015-07-10 14:50
Machine Schedule
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
ZOJ 1364 Machine Schedule(二分图最大匹配)的更多相关文章
- 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 ...
- UVA1194 Machine Schedule[二分图最小点覆盖]
题意翻译 有两台机器 A,B 分别有 n,m 种模式. 现在有 k 个任务.对于每个任务 i ,给定两个整数$ a_i\(和\) b_i$,表示如果该任务在 A上执行,需要设置模式为 \(a_i\): ...
- HDU 1150 Machine Schedule (二分图最小点覆盖)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1150 有两个机器a和b,分别有n个模式和m个模式.下面有k个任务,每个任务需要a的一个模式或者b的一个 ...
- 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, ...
- [poj1325] Machine Schedule (二分图最小点覆盖)
传送门 Description As we all know, machine scheduling is a very classical problem in computer science a ...
- POJ-1325 Machine Schedule 二分图匹配 最小点覆盖问题
POJ-1325 题意: 有两台机器A,B,分别有n,m种模式,初始都在0模式,现在有k项任务,每项任务要求A或者B调到对应的模式才能完成.问最少要给机器A,B调多少次模式可以完成任务. 思路: 相当 ...
- poj 2724 Purifying Machine(二分图最大匹配)
题意: 有2^N块奶酪,编号为00...0到11..1. 有一台机器,有N个开关.每个开关可以置0或置1,或者置*.但是规定N个开关中最多只能有一个开关置*. 一旦打开机器的开关,机器将根据N个开关的 ...
- (step6.3.3)hdu 1150(Machine Schedule——二分图的最小点覆盖数)
题目大意:第一行输入3个整数n,m,k.分别表示女生数(A机器数),男生数(B机器数),以及它们之间可能的组合(任务数). 在接下来的k行中,每行有3个整数c,a,b.表示任务c可以有机器A的a状态或 ...
- POJ - 1325 Machine Schedule 二分图 最小点覆盖
题目大意:有两个机器,A机器有n种工作模式,B机器有m种工作模式,刚開始两个机器都是0模式.假设要切换模式的话,机器就必须的重新启动 有k个任务,每一个任务都能够交给A机器的i模式或者B机器的j模式完 ...
随机推荐
- 【LCA倍增】POJ1330-Nearest Common Ancestors
[知识点:离线算法&在线算法] 一个离线算法,在开始时就需要知道问题的所有输入数据,而且在解决一个问题后就要立即输出结果. 一个在线算法是指它可以以序列化的方式一个个的处理输入,也就是说在开始 ...
- svn的安装(整合apache、ldap)包括错误解决post commit FS processing had error
2013年12月5日 admin 发表评论 阅读评论 以下是centos环境下,以yum安装apache及其相关软件.svn使用源码包编译,使用官网最新的1.8.5版本. 一.安装apache ope ...
- IE刷新后,文本框的值不变
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- http://blog.csdn.net/jhg1204/article/details/45013987
http://blog.csdn.net/jhg1204/article/details/45013987
- Mac上安装使用Nginx
1.brew search nginx 2.brew install nginx 启动nginx ,sudo nginx ;访问localhost:8080 发现已出现nginx的欢迎页面了. 备注: ...
- C# 多线程控制 通讯
一.多线程的概念 Windows是一个多任务的系统,如果你使用的是windows 2000及其以上版本,你可以通过任务管理器查看当前系统运行的程序和进程.什么是进程呢?当一个程序开始运行时,它就是一 ...
- Android中的Service组件具体解释
Service与Activity的差别在于:Service一直在后台执行,他没实用户界面,绝不会到前台来. 一,创建和配置Service 开发Service须要两个步骤:1.继承Service子类,2 ...
- 使用Python发送电子邮件
使用python发送邮件并不难,这里使用的是SMTP协议. Python标准库中内置了smtplib,使用它发送邮件只需提供邮件内容与发送者的凭证即可. 代码如下: # coding:utf-8 im ...
- ant design pro 表单
1.Input Enter事件 <input onKeyUp={this.onKeyUp} onPressEnter={this.enter} /> onKeyUp = (e) => ...
- ant-design 实现 添加页面
1.逻辑代码 /** * 添加用户 */ import React,{PureComponent} from 'react' import {Card,Form,Input,Select,Button ...