二分图最小覆盖点数（最大匹配）：Machine Schedule

Machine ScheduleTime 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 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30

Sample Output

3

这个题实际上就是求二分图的最小覆盖点数，用最少的点数覆盖尽量多的边

每条边代表一个工作，每多用一点就代表重启一次机器

另外要注意的是机子开始时0状态，所以所有可以在0状态情况搞定的任务不算进要考虑的边中

公式是最小覆盖点数 = 最大匹配，这个有个定理，证明见：点击打开链接

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>using namespace std;const int MAXN=1010;int uN,vN;  //u,v数目int g[MAXN][MAXN];//编号是0~n-1的int linker[MAXN];//是否有父亲，如果有，父亲是谁bool used[MAXN]; //在本次dfs中是否在路径中被指过bool dfs(int u){    int v;    for(v=0;v<vN;v++)        if(g[u][v]&&!used[v]){            used[v]=true;            if(linker[v]==-1||dfs(linker[v])){//v没有父亲或v的父亲有其他儿子                linker[v]=u;                return true;            }        }    return false;}int hungary(){    int res=0;    int u;    memset(linker,-1,sizeof(linker));    for(u=0;u<uN;u++){        memset(used,0,sizeof(used));        if(dfs(u))  res++;    }    return res;}int main(){    int n, m, k;    int i;    while(cin >> n && n){        scanf("%d %d", &m, &k);        memset(g, 0, sizeof(g));        uN = n;        vN = m;        for(i = 0; i < k; i ++){            int x, y, z;            scanf("%d %d %d",&z, &x, &y);            if(x * y)                g[x][y] = 1;        }        printf("%d\n", hungary());    }    return 0;}