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?pid=1150
Description
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 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
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
HINT
题意
有2个机器m个任务,每一个任务在a机器需要状态x,在b机器需要状态y,然后每个机器开始状态都是0,改变状态花费为1,然后问你最小花费完成这些任务
题解:
把每一个任务都当成一个边,把x状态和y连接起来,然后就是求最小的点来覆盖所有的边,就是一个最少点覆盖问题
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/* */
//************************************************************************************** inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int ma[maxn][maxn];
int vis[maxn];
int match[maxn];
int n,m;
vector<int> e[maxn];
int dfs(int a)
{
for(int i=;i<e[a].size();i++)
{
if(vis[e[a][i]]==)
{
vis[e[a][i]]=;
if(match[e[a][i]]==-||dfs(match[e[a][i]]))
{
match[e[a][i]]=a;
return ;
}
}
}
return ;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==)
break;
memset(match,-,sizeof(match));
for(int i=;i<n;i++)
e[i].clear();
m=read();
int k=read();
for(int i=;i<k;i++)
{
int a=read();
int x=read(),y=read();
if(x>&&y>)
{
e[x].push_back(y);
}
}
int ans=;
for(int i=;i<n;i++)
{
memset(vis,,sizeof(vis));
if(dfs(i)==)
ans++;
}
printf("%d\n",ans);
} }
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