poj 1325 最小顶点覆盖

Machine Schedule

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12559 Accepted: 5368

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

Source

Beijing 2002

每个任务存在一个(xi,yi)的关系，要么被机器A的xi模式完成，要么被机器B的yi模式完成。将两个机器的n、m个模式作为图中点，对于每个任务从点xi到yi连接一条边。

即任务作为边，并且每个任务被边的两个端点其中一个覆盖即可，所以这题就转化为最小顶点覆盖模型。 |最小顶点覆盖|=|最大匹配|，这题的图是二分图，只需要用匈牙利求最大匹配就可以了。

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <vector>#include <set>#include <map>#include <stack>#include <queue>using namespace std;vector<int> g[205];int match[205];int vis[205];int n,m,k;int dfs(int u){    vis[u]=1;    for(int i=0; i<g[u].size(); i++){        int v=g[u][i], mt=match[v];        if(mt==-1 || !vis[mt]&&dfs(mt)){            match[v]=u;            match[u]=v;            return 1;        }    }    return 0;}int hungry(){    memset(match, -1, sizeof(match));    int res=0;    for(int i=1; i<=n; i++)        if(match[i]==-1){            memset(vis, 0, sizeof(vis));            res+=dfs(i);        }    return res;}int main(){    while(scanf("%d", &n)==1 && n){        scanf("%d%d", &m, &k);        for(int i=0; i<=n+m; i++)g[i].clear();        int t, a,b;        for(int i=1; i<=k; i++){            scanf("%d%d%d", &t, &a, &b);            if(a==0||b==0) continue;            g[a].push_back(b+n);            g[n+b].push_back(a);        }        cout<<hungry()<<endl;    }    return 0;}