usaco 4.2 Job Processing(贪心)

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文章标签：

#processing #each #jobs #output #任务调度 #performance

本文探讨了一种生产流水线任务调度问题，通过算法优化找到完成所有任务所需的时间。详细介绍了输入输出格式，提供了示例输入输出，并通过代码实现了解决方案。

Job Processing
IOI'96

A factory is running a production line that requires two operations to be performed on each job: first operation "A" then operation "B". Only a certain number of machines are capable of performing each operation.

Figure 1 shows the organization of the production line that works as follows. A type "A" machine takes a job from the input container, performs operation "A" and puts the job into the intermediate container. A type "B" machine takes a job from the intermediate container, performs operation "B" and puts the job into the output container. All machines can work in parallel and independently of each other, and the size of each container is unlimited. The machines have different performance characteristics, a given machine requires a given processing time for its operation.

Give the earliest time operation "A" can be completed for all N jobs provided that the jobs are available at time 0. Compute the minimal amount of time that is necessary to perform both operations (successively, of course) on all N jobs.

PROGRAM NAME: job

INPUT FORMAT

Line 1:	Three space-separated integers: N, the number of jobs (1<=N<=1000). M1, the number of type "A" machines (1<=M1<=30) M2, the number of type "B" machines (1<=M2<=30)
Line 2..etc:	M1 integers that are the job processing times of each type "A" machine (1..20) followed by M2 integers, the job processing times of each type "B" machine (1..20).

SAMPLE INPUT (file job.in)

5 2 3
1 1 3 1 4

OUTPUT FORMAT

A single line containing two integers: the minimum time to perform all "A" tasks and the minimum time to perform all "B" tasks (which require "A" tasks, of course).

SAMPLE OUTPUT (file job.out)

3 5

题目： http://ace.delos.com/usacoprob2?a=6usYpMytQ3l&S=job

题意：给你n个任务，每个任务要用ab两个步骤完成，有m1台机器做a步骤，m2台机器做b步骤，求完成a步骤的时间，和完成b步骤的时间

分析：这题显然是任务调度问题，对于第一问，也就是完成a步骤，对于第二问，一开始我以为是网络流问题，因为这道题放在这个地方= =，其实还是个贪心问题，我们只要假设没有a步骤，然后同a步骤的做法，求出每个任务完成的具体时间，ab两个时间时间表反向相加的最大值就是第二问

代码：

/*
ID: 15114582
PROG: job
LANG: C++
*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int a[33],b[33],s1[33],s2[33],aa[1111],bb[1111];
int main()
{
    freopen("job.in","r",stdin);
    freopen("job.out","w",stdout);
    int i,j,k,n,m1,m2,t;
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
        for(i=0;i<m1;++i)
            scanf("%d",&a[i]),s1[i]=0;;
        for(i=0;i<m2;++i)
            scanf("%d",&b[i]),s2[i]=0;
        for(t=0;t<n;++t)
        {
            k=1e9;
            for(i=0;i<m1;++i)
                if(s1[i]+a[i]<k)
                {
                    k=s1[i]+a[i];
                    j=i;
                }
            aa[t]=s1[j]=k;
            k=1e9;
            for(i=0;i<m2;++i)
                if(s2[i]+b[i]<k)
                {
                    k=s2[i]+b[i];
                    j=i;
                }
            bb[t]=s2[j]=k;
        }
        sort(aa,aa+n);
        sort(bb,bb+n);
        printf("%d ",aa[n-1]);
        for(k=i=0;i<n;++i)
            k=max(k,aa[i]+bb[n-i-1]);
        printf("%d\n",k);
    }
    return 0;
}