HDU 1150 二分图

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

 

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

 

有k 个任务可以在 A 机器的某一种模式或者 B 机器上的某一种模式运行，

A 机器有 n 个模式， B 机器有 m 个模式，

对于每一台机器不同模式之间需要切换，

问完成 k 个任务需要至少切换多少次

最小顶点覆盖= 最大匹配数...直接套模板

#include  <iostream>
using namespace std;
const int maxn=555;
int map[maxn][maxn];
int vis[maxn];//是否访问过
int link[maxn];//记录 y中的点i 在x中所匹配的点x的编号 
int Max;//最大匹配数
int x,y; //矩阵的行列
bool dfs(int x)
{
    for(int i=1;i<=y;i++)
    {
        if(map[x][i]&&!vis[i])//x到i相邻(有边) 且 节点i未被搜索  
        {
            vis[i]=1;//标记节点i已被搜索
            if(link[i]==-1||dfs(link[i]))//link[i]==0 : 如果i不属于前一个匹配M  
            {//dfs(link[i]) : 如果被i匹配到的节点可以寻找到增广路 
                link[i]=x;//那么可以更新匹配
                return true;//返回匹配成功的标志
            }
        }
    }
    return false;
}
void hungary()
{
    for(int i=1;i<=x;i++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(i))Max++;
    }
}
int main()
{
    int k,p,a,b,i;
    while(scanf("%d",&x)&&x)
    {
        scanf("%d%d",&y,&k);
        memset(map,0,sizeof(map));
        memset(link,-1,sizeof(link));
        for(i=1;i<=k;i++)
        {
            scanf("%d%d%d",&p,&a,&b);
            map[a][b]=1;
        }
        Max=0;
        hungary();
        printf("%d\n",Max);
    }
    return 0;
}