Machine Schedule

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1150

Description

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

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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';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=0;i<e[a].size();i++)
    {
        if(vis[e[a][i]]==0)
        {
            vis[e[a][i]]=1;
            if(match[e[a][i]]==-1||dfs(match[e[a][i]]))
            {
                match[e[a][i]]=a;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        memset(match,-1,sizeof(match));
        for(int i=0;i<n;i++)
            e[i].clear();
        m=read();
        int k=read();
        for(int i=0;i<k;i++)
        {
            int a=read();
            int x=read(),y=read();
            if(x>0&&y>0)
            {
                e[x].push_back(y);
            }
        }
        int ans=0;
        for(int i=1;i<n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(dfs(i)==1)
                ans++;
        }
        printf("%d\n",ans);
    }

}