HDU——T 1150 Machine Schedule
http://acm.hdu.edu.cn/showproblem.php?pid=1150
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9463 Accepted Submission(s): 4757
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.
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 <cstring>
#include <cstdio> using namespace std; int n,m,t,match[];
bool map[][],vis[]; bool find(int u)
{
for(int v=;v<=m;v++)
if(map[u][v]&&!vis[v])
{
vis[v]=;
if(!match[v]||find(match[v]))
{
match[v]=u;
return true;
}
}
return false;
} int main()
{
for(int ans=;scanf("%d",&n)&&n;ans=)
{
scanf("%d%d",&m,&t);
for(int x,y,z;t--;map[y][z]=)
scanf("%d%d%d",&x,&y,&z);
for(int i=;i<=n;i++)
{
if(find(i)) ans++;
memset(vis,,sizeof(vis));
}
printf("%d\n",ans);
memset(map,,sizeof(map));
memset(match,,sizeof(match));
}
return ;
}
HDU——T 1150 Machine Schedule的更多相关文章
- 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(最小顶点覆盖)
pid=1150">Machine Schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/327 ...
- 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 ...
- 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) ...
随机推荐
- ansible 连通测试
[root@ftp:/root] > ansible ansible01 -m ping ansible01 | UNREACHABLE! => { "changed" ...
- NOIP2018提高组省一冲奖班模测训练(四)
NOIP2018提高组省一冲奖班模测训练(四) 这次比赛只AC了第一题,而且花了40多分钟,貌似是A掉第一题里面最晚的 而且还有一个半小时我就放弃了…… 下次即使想不出也要坚持到最后 第二题没思路 第 ...
- Vue-router入门
1.npm install vue-router --save-dev 安装路由包,在安装脚手架时实际上可以直接安装 2.解读核心文件 router/index.js文件 import Vue fro ...
- 机器学习关于AUC的理解整理
AUC 几何意义:ROC曲线与X轴的面积 https://blog.csdn.net/luo3300612/article/details/80367901 AUC物理意义:随机给定一个正样本和一个负 ...
- java--web学习总结<转>
http://www.cnblogs.com/xdp-gacl/p/3729033.html
- 2015 Multi-University Training Contest 5 hdu 5352 MZL's City
MZL's City Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total ...
- server.htaccess 具体解释以及 .htaccess 參数说明
.htaccess文件(或者"分布式配置文件")提供了针对文件夹改变配置的方法. 即.在一个特定的文档文件夹中放置一个包括一个或多个指令的文件, 以作用于此文件夹及其所有子文件夹. ...
- android学习笔记(5)Activity生命周期学习
每一个activity都有它的生命周期,开启它,关闭它,跳转到其他activity等等,都会自己主动调用下面某种方法.对这些个方法覆写后观察学习. protected void onCreate(Bu ...
- HDU5411CRB and Puzzle(矩阵高速幂)
题目链接:传送门 题意: 一个图有n个顶点.已知邻接矩阵.问点能够反复用长度小于m的路径有多少. 分析: 首先我们知道了邻接矩阵A.那么A^k代表的就是长度为k的路径有多少个. 那么结果就是A^0+A ...
- 广播BroadcastReceiver(2)
有序广播的优先级: 发送有序广播的方法有: public void sendOrderedBroadcast(Intent intent,String receiverPermis ...