HDU 1150

HDU - 1150

Machine Schedule

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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k个任务，可用A机器或者B机器执行，A机器有n种模式，B机器有m种模式，每个任务可用A或B机器的一种模式执行，每次切换模式代价为1，求执行完所有任务的最小的代价。
思路：把A机器的所有模式看成X集合，把B机器的所有模式看成Y集合，若同时可做一种任务， 则连边。这样，边就代表k个任务。问题就转化为求最小的点集，能够覆盖所有的边 = =
最小顶点覆盖数 = 最大匹配数  求最大匹配。
AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, m,k;int g[105][105], linker[105], vis[105];int Find(int u){    for(int i = 1;i < m;i ++)    {        if(!vis[i] && g[u][i])        {            vis[i] = 1;            if(linker[i] == -1 || Find(linker[i]))            {                linker[i] = u;                return 1;            }        }    }    return 0;}int main(){    while(scanf("%d", &n) != EOF && n)    {        scanf("%d%d", &m, &k);        memset(g, 0, sizeof(g));        memset(linker, -1, sizeof(linker));        while(k --)        {            int id, u, v;            scanf("%d%d%d", &id, &u, &v);            g[u][v] = 1;        }        int ans  = 0;        for(int i = 1;i < n;i ++)        {            memset(vis, 0, sizeof(vis));            if(Find(i))                ans ++;        }        printf("%d\n", ans);    }    return 0;}