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  • POJ1325 Machine Schedule

    Machine Schedule
    Time Limit: 1000MSMemory Limit: 10000K
    Total Submissions: 17934Accepted: 7483

    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

    有两个机器A和B,A机器有n个模式,B机器有m个模式,两个机器最初在0模式。然后有k个作业,每个作业有三个参数i,a,b。i代表作业编号,a和b代表第i作业要么在A机器的a模式下完成或者在B机器的b模式下完成。问两个机器总共最少变换多少次可以完成所有作业

    题解

    把机器A和机器B的模式拆成节点
    2要素:每个任务的x,y必选其一
    所以这是二分图最小点覆盖问题,根据Konig定理转化为最大匹配。

    #include<iostream>
    #include<vector>
    #include<cstring>
    #define rg register
    #define il inline
    #define co const
    template<class T>il T read(){
        rg T data=0,w=1;rg char ch=getchar();
        for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
        for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
        return data*w;
    }
    template<class T>il T read(rg T&x) {return x=read<T>();}
    typedef long long ll;
    using namespace std;
    
    co int N=101;
    int n,m,k,f[N],ans;
    bool v[N];
    vector<int> e[N];
    
    bool dfs(int x){
    	for(unsigned i=0;i<e[x].size();++i){
    		int y=e[x][i];
    		if(v[y]) continue;
    		v[y]=1;
    		if(!f[y]||dfs(f[y])){
    			f[y]=x;
    			return 1;
    		}
    	}
    	return 0;
    }
    void Machine_Schedule(){
    	read(m),read(k);
    	for(int i=1;i<n;++i) e[i].clear();
    	for(int i=0,x,y;i<k;++i){
    		read<int>(),read(x),read(y);
    		if(x&&y) e[x].push_back(y);
    	}
    	memset(f,0,sizeof f);
    	ans=0;
    	for(int i=1;i<n;++i){
    		memset(v,0,sizeof v);
    		ans+=dfs(i);
    	}
    	printf("%d
    ",ans);
    }
    int main(){
    	while(read(n)) Machine_Schedule();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/autoint/p/10972227.html
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