HDU1150：Machine Schedule（最小点覆盖）

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9117    Accepted Submission(s): 4575

Problem 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 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30

 

Sample Output

3

 

Source

Asia 2002, Beijing (Mainland China)

题意：有A，B两台机器，K个任务，每个任务可以由A或B中某个模式完成，两机器初始模式均为0，而机器切换模式需要重启，问最少重启几次能完成任务。

思路：建二分图，任务对应的两种模式建边，那么显然计算选最少的点，使所有边被覆盖，这就是最小点覆盖数的定义，同时最小点覆盖数=最大匹配数，匈牙利算法即可。

# include <iostream># include <cstdio># include <cstring>using namespace std;int n, m, k, Map[103][203], vis[203], link[203];bool dfs(int u){    for(int i=n+1; i<n+m; ++i)    {        if(!vis[i] && Map[u][i])        {            vis[i] = 1;            if(link[i]==-1 || dfs(link[i]))            {                link[i] = u;                return true;            }        }    }    return false;}int main(){    int a, b, c;    while(~scanf("%d",&n),n)    {        scanf("%d%d",&m,&k);        memset(Map, 0, sizeof(Map));        memset(link, -1, sizeof(link));        while(k--)        {            scanf("%d%d%d",&a,&b,&c);            if(b && c) Map[b][c+n] = 1;        }        int ans = 0;        for(int i=1; i<n; ++i)        {            memset(vis, 0, sizeof(vis));            if(dfs(i)) ++ans;        }        printf("%d\n",ans);    }    return 0;}