传送门

描述

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.

输入

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.

输出

对输入的每组数据，按照如下格式输出：

The output should be one integer per line, which means the minimal times of restarting machine.

样例

• 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
• Output
  3

题解

• 题意：现在机器A有n种状态（1~n)，机器B有m种状态(1~m)，k项工作。每项工作可以让机器A的状态x做，也可以让机器B的状态y做。每台机器某一时刻只能处于某一种状态，显然，要完成所有工作就需要不断改变两台机器的状态。两台机器的初始状态都是0，给出k项工作的x，y，k项工作完成顺序不限，求出想要完成所有工作，A、B两机器需要改变状态次数和最小是多少。
• 根据每项工作i，将xi、yi连接起来，构成二分图，这条连线就表示工作i。我们需要找到最少的点使得所有的边被覆盖，也就是最小点覆盖。最小点覆盖=二分图最大匹配。

Code

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#include
         
           #include
          
            #define INIT(a,b) memset(a,b,sizeof(a)) #define LL long long using namespace std; const int inf=0x3f3f3f3f; const int maxn=1e2+7; const int mod=1e9+7; int n,m,t; int line[maxn][maxn],used[maxn],another[maxn]; bool Find(int x){
            for(int i=1;i<=m;i++){
                if(!used[i]&&line[x][i]){
                    used[i]=1;             if(!another[i]||Find(another[i])){
                        another[i]=x;                 return true;             }         }     }     return false; } int Count(){
            int ans=0;     for(int i=1;i<=n;i++){
                INIT(used,0);         ans+=Find(i);     }     return ans; } int main(){
            int a,b,c;     while(~scanf("%d",&n)&&n){
                scanf("%d%d",&m,&t);         INIT(line,0);INIT(another,0);         while(t--){
                    scanf("%d%d%d",&a,&b,&c);             line[b][c]=1;         }         printf("%d\n",Count());     } 	return 0; }