HDOJ 1150 Machine Schedule
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
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.
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
#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
- 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) ...
- 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 ...
- 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/ ...
- 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 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
Machine Schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- 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
- 51nod1730 涂边
题目描述 题解 八级sb题 显然可以想到状压 枚举当前的宽度\(I\),设\(f[s]\)表示在当前的宽度下选的竖边的状态为s 再设\(g[s1][s2]\)表示状态s1转移到s2的方案数,枚举中间横 ...
- 阿里云 Serverless 应用引擎(SAE)发布 v1.2.0,支持一键启停、NAS 存储、小规格实例等实用特性
近日,阿里云 Serverless 应用引擎(SAE)发布 v1.2.0版本,新版本实现了以下新功能/新特性: 一键启停开发测试环境:企业开发测试环境一般晚上不常用,长期保有应用实例,闲置浪费很高.使 ...
- 拆系数$FFT$($4$遍$DFT$)
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> ...
- [洛谷P3940]:分组(贪心+并查集)
题目传送门 题目描述 小$C$在了解了她所需要的信息之后,让兔子们调整到了恰当的位置.小$C$准备给兔子们分成若干个小组来喂恰当的胡萝卜给兔子们吃.此时,$n$只兔子按一定顺序排成一排,第$i$只兔子 ...
- CSS基础-background的那些属性
background的那些属性 background:背景的意思常用的六个属性 1.background-color:背景颜色 2.background-image:背景图像 3.background ...
- Linux超全实用指令大全
参考 Linux超全实用指令大全
- join当前线程等待指定的线程结束后才能继续运行
模拟一个QQ游戏大厅斗地主 /** sleep(休眠.睡眠) join当前线程等待指定的线程结束后才能继续运行 */ class Player extends Thread{ private Stri ...
- windows系统下,在C#程序中自动安装字体
在Windows系统中,原有自带的字体样式有限,有时候我们的程序会使用到个别稀有或系统不自带的字体.因此我们需要将字体打包到程序中,当程序启动时,检测系统是否有该字体,如果没有则安装该字体,也可以动态 ...
- SQLite入门语句之约束
一.SQLite约束之NOT NULL 确保某列不能有 NULL 值.默认情况下,列可以保存 NULL 值.如果您不想某列有 NULL 值,那么需要在该列上定义此约束,指定在该列上不允许 NULL 值 ...
- Powershell&TFS_Part 1
目录 目录 前言 TFS 对象模型 Powershell Powershell面向对象 Powershell默认会在PC中设置执行脚本权限 调试脚本 断点 Step Microsoft Visual ...