POJ 1325、ZOJ 1364、HDU 1150 Machine Schedule

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

Sample Output

3

       题目大意：有Ａ，Ｂ两种机器，给你三个数ｎ，ｍ，ｋ，分别表示机器Ａ有ｎ中工作模式（编号０　～　ｎ－１），机器Ｂ有ｍ种工作模式（编号０～ｍ－１），共有ｋ个任务，每种任务均可以在机器Ａ，Ｂ的一个模式下完成。接下来输入ｋ行，每行三个整数ａ，ｂ，ｃ，其中，ａ为任务编号，ｂ表示该任务可在机器Ａ的第ｂ种模式下完成，ｃ表示该任务可在机器Ｂ的第ｃ中模式下完成。但机器Ａ，Ｂ在变换模式时均需重启，让你完成所有的任务并使机器重启的次数最小。（机器Ａ，Ｂ初始时均在第０模式）。
   

    　　解题思路：此题是求二分图的最小点覆盖。有以下定理：二分图的点覆盖数　＝　匹配数。　建图：把A的n种模式和B的m种模式看做顶点，如果某人可在A的第i个模式和B的第j个模式下完成，则将顶点Ai 和 Bj 之间连一条边。
   

       请看代码：
   

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std ;
const int MAXN = 105 ;
short g[MAXN][MAXN] ;
bool vis[MAXN] ;
short cx[MAXN] , cy[MAXN] ;
int n , m , k ;
void init()
{
    memset(g , 0 , sizeof(g)) ;
    memset(cx , -1 , sizeof(cx)) ;
    memset(cy , -1 , sizeof(cy)) ;
    int i ;
    for(i = 0 ; i < k ; i ++)
    {
        int a , b , c ;
        scanf("%d%d%d" , &a , &b ,&c) ;
        if(b != 0 && c != 0)
        {
            g[b][c] = 1 ;
        }
    }
}
int path(int v)
{
    int i ;
    for(i = 0 ; i < m ; i ++)
    {
        if(g[v][i] && !vis[i])
        {
            vis[i] = 1 ;
            if(cy[i] == -1 || path(cy[i]))
            {
                cy[i] = v ;
                cx[v] = i ;
                return 1 ;
            }
        }
    }
    return 0 ;
}
void solve()
{
    int i ;
    int ans = 0 ;
    for(i = 0 ; i < n ; i ++)
    {
        if(cx[i] == -1)
        {
            memset(vis , 0 , sizeof(vis)) ;
            if(path(i))
            ans ++ ;
        }
    }
    printf("%d
" , ans) ;
}
int main()
{
    while (scanf("%d" , &n) != EOF)
    {
        if(n == 0)
        break ;
        scanf("%d%d" , &m , &k) ;
        init() ;
        solve() ;
    }
    return 0 ;
}