POJ 1325 二分图匹配/匈牙利算法

最新推荐文章于 2022-07-21 19:42:41 发布

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原文链接：http://www.cnblogs.com/codeyuan/p/4254435.html

本文介绍了一种使用二分图和匈牙利算法解决两台机器调度问题的方法，旨在最小化机器重启次数，具体通过输入配置解析约束，输出最优的机器操作序列。

Machine Schedule
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 11922 Accepted: 5077
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
Source

Beijing 2002

<span style="color:#000099;">/****************************************
     author    : Grant Yuan
     time      : 2014/9/17/ 00:06
     algorithm : 二分图/匈牙利算法
     source    : POJ 1325
****************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#define MAX 1007

using namespace std;
vector<int> G[MAX];
bool used[MAX];
int match[MAX];
int n,m,k;

bool dfs(int v)
{
    for(int i=0;i<G[v].size();i++)
    {
        int u;
        u=G[v][i];
        if(!used[u]){
            used[u]=1;
            if(match[u]==-1||dfs(match[u]))
            {
                match[u]=v;
                return true;
            }
        }
    }
    return false;
}

int bipartite_matching()
{
    int res=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i)) res++;
    }
    return res;
}

int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=0;i<MAX;i++)
            while(!G[i].empty()) G[i].pop_back();
        scanf("%d%d",&m,&k);
        for(int i=0;i<k;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(b!=0&&c!=0) G[b].push_back(c);
        }
        int ans=bipartite_matching();
        printf("%d\n",ans);
    }
    return 0;
}
</span>

转载于:https://www.cnblogs.com/codeyuan/p/4254435.html