2015年ACM训练二分图（2）：Machine Schedule

                                                            C - Machine Schedule

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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

 题意：有两台机器A和B以及N个需要运行的任务。每台机器有M种不同的模式，而每个任务都恰好在一台机器上运行。如果它在机器A上运行，则机器A需要设置为模式xi,如果它在机器B上运行，则机器A需要设置为模式yi。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。初始时是模式0.

   

最大匹配数：最大匹配的匹配边的数目

最小点覆盖数：选取最少的点，使任意一条边至少有一个端点被选择

最大独立数：选取最多的点，使任意所选两点均不相连

最小路径覆盖数：对于一个 DAG（有向无环图），选取最少条路径，使得每个顶点属于且仅属于一条路径。路径长可以为 0（即单个点）。

定理1：最大匹配数 = 最小点覆盖数（这是 Konig 定理）

定理2：最大匹配数 = 最大独立数

定理3：最小路径覆盖数 = 顶点数 - 最大匹配数

代码如下：

#include <iostream>
#include <cstring>
using namespace std;
int line[510][510],boy[510],used[510];
int n,m;
int find(int x)
{
    int i;
    for(i=1; i<=m; i++)
    {
        if(line[x][i]==1&&used[i]==0)
        {
            used[i]=1;
            if(boy[i]==0||find(boy[i]))
            {
                boy[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,k,x,y,sum,q;
    while(cin>>n)
    {
        if(n==0)
        {
            break;
        }
        else
        {
            cin>>m>>k;
            memset(line,0,sizeof(line));
            memset(boy,0,sizeof(boy));
            memset(used,0,sizeof(used));
            for(i=0; i<k; i++)
            {
                cin>>q>>x>>y;
                line[x][y]=1;
            }
            sum=0;
            for(i=1; i<=n; i++)
            {
                memset(used,0,sizeof(used));
                if(find(i))
                {
                    sum++;
                }
            }
            cout<<sum<<endl;
        }

    }
    return 0;
}