HDU

Machine Schedule

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

题目大意：有两台机器A和B，A机器有n种工作方式，B机器有m种工作方式。共有k个任务。每个任务恰好在一条机器上运行。

如果任务在A机器上运行，就需要转换为模式X_i，如果在B机器上运行，就需要转换为模式Y_i。

每台机器上的任务可以按照任意顺序执行，但是每台机器每转换一次模式需要重启一次。

请合理为每个任务安排一台机器并合理安排顺序，使得机器重启次数尽量少。

 

解题思路：

把机器A的N种模式作为二分图的左部，机器B的M种模式作为二分图的右部，如果某个任务可以使用机器A的模式xi也可以使用机器B的模式yi完成，则连接xi,yi。

题目要求使机器重启的次数要尽量少，又要把所有的任务都执行完，也就可以把题目转换成最小顶点覆盖，根据二分图的性质：最小顶点覆盖=最大匹配数。

 

cb学长博客匈牙利算法讲解https://blog.csdn.net/cillyb/article/details/55511666

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<set>
#include<stack>
#include<algorithm>
#include<vector>
#include<iterator>
#define MAX 1005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;

typedef long long ll;
int n,m;
int line[MAX][MAX],used[MAX],mm[MAX];
int find(int x){  
    int i,j;  
    for(j=1;j<=m;j++){
        if(line[x][j]&&!used[j]){
            used[j]=1;
            if(!mm[j]||find(mm[j])){
                mm[j]=x;
                return 1;
            }
        }
    }
    return 0;  
}  
int main()
{
    int xx,x,y,k,i;
    while(scanf("%d",&n)&&n>0){
        scanf("%d%d",&m,&k);
        memset(line,0,sizeof(line));
        memset(mm,0,sizeof(mm));
        for(i=1;i<=k;i++){
            scanf("%d%d%d",&xx,&x,&y);
            if(x>0&&y>0){
                line[x][y]=1;
            }
        }
        int c=0;
        for(i=1;i<=n;i++){
            memset(used,0,sizeof(used));
            if(find(i)) c++;
        }
        printf("%d
",c);
    }
    return 0;
}