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)
Total Submission(s): 6701 Accepted Submission(s): 3358
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.
#include<stdio.h>
#include<string.h>
#define MAX 1100
int map[MAX][MAX],vis[MAX],mode[MAX];
int t,n,m;
int find(int x)
{
int i,j;
for(i=0;i<m;i++)
{
if(map[x][i]&&vis[i]==0)
{
vis[i]=1;
if(mode[i]==0||find(mode[i]))
{
mode[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int j,i,s,k;
int x,y;
while(scanf("%d",&n),n)
{
memset(map,0,sizeof(map));
memset(mode,0,sizeof(mode));
scanf("%d%d",&m,&k);
while(k--)
{
scanf("%d%d%d",&t,&x,&y);
if(x&&y)
map[x][y]=1;
}
s=0;
for(i=0;i<n;i++)
{
memset(vis,0,sizeof(vis));
if(find(i))
s++;
}
printf("%d\n",s);
}
return 0;
}
hdoj 1150 Machine Schedule【匈牙利算法+最小顶点覆盖】的更多相关文章
- HDU - 1150 POJ - 1325 Machine Schedule 匈牙利算法(最小点覆盖)
Machine Schedule As we all know, machine scheduling is a very classical problem in computer science ...
- POJ 1325 && 1274:Machine Schedule 匈牙利算法模板题
Machine Schedule Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12976 Accepted: 5529 ...
- POJ1325 Machine Schedule 【二分图最小顶点覆盖】
Machine Schedule Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11958 Accepted: 5094 ...
- 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, ...
- 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/3 ...
- hdoj 1054 Strategic Game【匈牙利算法+最小顶点覆盖】
Strategic Game Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- (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 ...
- 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/ ...
随机推荐
- POJ 3371 Flesch Reading Ease 无聊恶心模拟题
题目:http://poj.org/problem?id=3371 无聊恶心题,还是不做的好,不但浪费时间而且学习英语. 不过为了做出点技术含量,写了个递归函数... 还有最后判断es,ed,le时只 ...
- 将十进制的颜色制转换成ARGB
将一个十进制的颜色值转换成具体的ARGB 格式,起初,这看起来有些难,一直找不到方法,在网上也找不到具体的资料,最后在同事的指导下成功完成的转换,现分享出来,供大家参考,具体转换方法如下: /// & ...
- leetcode 第二题Add Two Numbers java
链接:http://leetcode.com/onlinejudge Add Two Numbers You are given two linked lists representing two n ...
- CISCO2691的OSPF点对点密文测评测试
都差不多,粘一个文件就能说明问题了. Router#show run Building configuration... Current configuration : bytes ! version ...
- [topcoder]BestRoads
http://community.topcoder.com/stat?c=problem_statement&pm=10172&rd=13515 http://community.to ...
- 心愿:做一个精简版MFC
我觉得自己有能力看懂MFC的C++代码和总体流程,但是由于MFC混杂了太多的东西,比如OLE等等不必要的东西,又做了无数的ASSERT判断,影响整体流程的理解.因此我要做一个精简版的MFC,而且能用它 ...
- SectionIndexer中的getSectionForPosition()与getPositionForSection()
大家在做字母索引的时候常常会用到SectionIndexer这个类,里面有2个重要的方法 1. getSectionForPosition()通过该项的位置,获得所在分类组的索引号 2. getP ...
- 翻译文章“AST 模块:用 Python 修改 Python 代码”---!!注意ironpathyon未实现此功能
https://github.com/upsuper/blog/commit/0214fdd084c4adf2de2ed9912d644fb59ce13a1c +Title: [翻译] AST 模块: ...
- 【HDOJ】3726 Graph and Queries
Treap的基础题目,Treap是个挺不错的数据结构. /* */ #include <iostream> #include <string> #include <map ...
- (转载)mysql分屏显示结果
(转载)http://blog.csdn.net/wylkeke/article/details/7280645 linux机器: 在mysql命令行输入pager more就可以分屏显示结果了,取消 ...