hdu 1150 二分匹配

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

Sample Output
3

题意：

给你两台机器A和B，A机器有n种模式，B机器有m种模式，初始时都是0，现在给你k个任务，每个任务可以由机器A的x模式完成或者机器B的y模式完成，而每次改变机器的模式都要重启一次，问你最少的重启次数使得完成所有任务！

题解：

其实就是建立二分图，求最小点覆盖，最小点覆盖就是求二分图最少的点，使得每条边至少和其中一个覆盖。最小点覆盖=最大匹配数。
问题转化为，求二分图中最少的点（模式）使得每个边（模式对应的任务）至少和其中的一个点覆盖
wrong answer了一次。
没考虑到题目：At the beginning they are both work at mode_0。
状态位0的情况不考虑，即不重启。

代码：

#include <bits/stdc++.h>using namespace std;int uN,vN,k;const int MAXN =1010;int g[MAXN][MAXN];int linker[MAXN];bool used[MAXN];bool dfs(int u){    for(int v=1;v<vN;v++)    {        if(g[u][v]&&!used[v])        {            used[v]=true;            if(linker[v]== -1||dfs(linker[v]))            {                linker[v]=u;                return true;            }        }    }    return false;}int hungary(){    int res = 0;    memset(linker,-1,sizeof(linker));    for(int u=1;u<uN;u++)    {        memset(used,false,sizeof(used));        if(dfs(u))        {            res++;        }    }    return res;}int main(){    while(cin>>uN&&uN)    {        memset(g,0,sizeof(g));         cin>>vN>>k;         int num,x,y;         for(int i=0;i<k;i++)         {             cin>>num>>x>>y;             g[x][y]=1;         }         cout<<hungary()<<endl;    }    return 0;}