浙大ZOJ 1012 Mainframe问题解决

一、工程代码及算法设计注释

--------------------------------------------------mainframe.h----------------------------------------------

#ifndef JLU_CCST_MAINFRAME_H#define JLU_CCST_MAINFRAME_H#include <vector>struct Job{int cpu,memory,arrive,deadline,reward,bonus,punishment;bool isExecuted;Job(int c=0,int m=0,int a=0,int d=0,int r=0,int b=0,int p=0,bool i=false):cpu(c),memory(m),arrive(a),deadline(d),reward(r),bonus(b),punishment(p),isExecuted(i) {} ;/**如果比job小，则返回-1；否则返回1.（由于本题中job的特性，任意2个job不会相等）*/int compare(const Job& job){if(arrive<job.arrive)return -1;if(arrive==job.arrive&&bonus<job.bonus)return -1;return 1;}};extern int caculateIncome(std::vector<Job>& jobList,int nCPU,int nMemory,int timeLineF);extern void testCaculateIncome();#endif//JLU_CCST_MAINFRAME_H

--------------------------------------------------mainframe.cpp----------------------------------------------

/**题目来源：浙大ACM在线测试——ZOJ，题目编号1012，题名"Mainframe"URL：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1012Author：hellogdc<[e-mail]>Finish Time:2014.01.10*//**问题分析：该题的关键是要将所有job排一列，然后从头开始逐个执行这些job。排序的指标有2个：主指标为到达时间，次指标为报酬。因为每个任务的报酬都不一样，所以任意2个job之间都有一个“先后”关系，所以这是一个全序关系。job的执行：由于CPU及内存的限制，所以job可能不会是按照排序好的队列顺序执行，因为排在前面的job可能因为CPU和内存限制导致执行不了，从而让位于后面的一些对资源要求较小的任务。算法设计：1. 基于任务到达时间和报酬，对所有任务进行排序，得到一个有序列表jobList.2. 如果jobList非空，则循环执行下面语句。2.1 设置当前时间：将队列最前面的job的到达时间设置为当前时间，同时设置资源：m=M,n=N。2.2 遍历jobList，找出合适的job进行执行。2.2.1 如果该队列元素的到达时间比当前时间大，则退出该循环，结束寻找。2.2.2 对于所有到达时间等于当前时间的job，做如下处理：如果当前的剩余资源能满足该job的要求，则该job将被执行，将其从队列中剔除，计算费用，更新当前剩余资源。否则，将其到达时间加1（这是为了能在下一小时被优先检测）.3. 统计那些Ui<=F但确未被执行的job，同时计算费用。*/#include <iostream>#include <vector>#include "mainframe.h"using namespace std;int caculateIncome(vector<Job>& jobList,int nCPU,int nMemory,int timeLineF){//1. 递减排序for(int i=0;i<jobList.size()-1;i++){for(int j=i+1;j<jobList.size();j++){if(jobList[i].compare(jobList[j])==-1){Job tmp=jobList[j];jobList[j]=jobList[i];jobList[i]=tmp;}}}//2. 计算那些被执行的job所产生的的费用int income=0;int head=0;while(true){//2.1 找到队列中尚未执行的“首席”jobwhile(head<jobList.size()){if(!jobList[head].isExecuted)break;head++;}if(head==jobList.size())break;int currentTime=jobList[head].arrive;if(currentTime>timeLineF-1)break;int nC=nCPU,nM=nMemory;bool isFound=false;//2.2 遍历jobList，找出合适的job进行执行。for(int i=head;i<jobList.size();i++){if(jobList[i].isExecuted)continue;//2.2.1 如果该队列元素的到达时间比当前时间大而且已到合适的job来执行，//则退出该循环，不再寻找。if(jobList[i].arrive>currentTime)break;//2.2.2 如果队列元素的到达时间等于当前时间，且当前的资源量能满足该job的要求，//则该job将被执行，将其从队列中剔除。计算费用。更新当前剩余资源。//否则，将其到达时间加1（这是为了能在下一小时被优先检测）.if( jobList[i].arrive=currentTime){if(jobList[i].cpu<=nC&&jobList[i].memory<=nM ){nC-=jobList[i].cpu;nM-=jobList[i].memory;isFound=true;jobList[i].isExecuted=true;income+=jobList[i].reward;if(currentTime+1<jobList[i].deadline)income+=( (jobList[i].deadline-currentTime-1)*jobList[i].bonus );elseincome-=( (currentTime+1-jobList[i].deadline)*jobList[i].punishment );}elsejobList[i].arrive++;//下次执行}}}//3. 统计那些Ui<=F但确未被执行的jobfor(int i=head;i<jobList.size();i++){if( (!jobList[i].isExecuted)&&jobList[i].deadline<=timeLineF )income-=( (timeLineF-jobList[i].deadline)*jobList[i].punishment );}return income;}/////////////////////////////Test Case///////////////////////////////////////void testCaculateIncome(){vector<Job> jobList;int cpu,memory,timelineF;//////////////////////////////cpu=4;memory=256;timelineF=10;jobList.push_back(Job(1,16,2,3,10,5,6));jobList.push_back(Job(2,128,2,4,30,10,5));jobList.push_back(Job(2,128,2,4,20,10,5));int income=caculateIncome(jobList,cpu,memory,timelineF);cout<<"Income="<<income<<endl;}

--------------------------------------------------main.cpp----------------------------------------------

#if 0#include "mainframe.h"int main(){testCaculateIncome();}#endif

二、提交给ZOJ的代码——算法核心代码

注：蛋疼的报了“Wrong Answer”错误。还未解决。

--------------------------------------------------submit_main.cpp----------------------------------------------

/**all codes are copied from mainframe.cpp*/#include <iostream>#include <vector>using namespace std;struct Job{int cpu,memory,arrive,deadline,reward,bonus,punishment;bool isExecuted;Job(int c=0,int m=0,int a=0,int d=0,int r=0,int b=0,int p=0,bool i=false):cpu(c),memory(m),arrive(a),deadline(d),reward(r),bonus(b),punishment(p),isExecuted(i) {} ;int compare(const Job& job){if(arrive<job.arrive)return -1;if(arrive==job.arrive&&bonus<job.bonus)return -1;return 1;}};int computeIncome(vector<Job>& jobList,int nCPU,int nMemory,int timeLineF){for(int i=0;i<jobList.size()-1;i++){for(int j=i+1;j<jobList.size();j++){if(jobList[i].compare(jobList[j])==-1){Job tmp=jobList[j];jobList[j]=jobList[i];jobList[i]=tmp;}}}int income=0;int head=0;while(true){while(head<jobList.size()){if(!jobList[head].isExecuted)break;head++;}if(head==jobList.size())break;int currentTime=jobList[head].arrive;if(currentTime>timeLineF-1)break;int nC=nCPU,nM=nMemory;bool isFound=false;for(int i=head;i<jobList.size();i++){if(jobList[i].isExecuted)continue;if(jobList[i].arrive>currentTime)break;if( jobList[i].arrive=currentTime){if(jobList[i].cpu<=nC&&jobList[i].memory<=nM ){nC-=jobList[i].cpu;nM-=jobList[i].memory;isFound=true;jobList[i].isExecuted=true;income+=jobList[i].reward;if(currentTime+1<jobList[i].deadline)income+=( (jobList[i].deadline-currentTime-1)*jobList[i].bonus );elseincome-=( (currentTime+1-jobList[i].deadline)*jobList[i].punishment );}elsejobList[i].arrive++;}}}for(int i=head;i<jobList.size();i++){if( (!jobList[i].isExecuted)&&jobList[i].deadline<=timeLineF )income-=( (timeLineF-jobList[i].deadline)*jobList[i].punishment );}return income;}int main(){int cpu,memory,timelineF;int no=1;int nJobs;while(cin>>timelineF){if(timelineF==0)break;if(no!=1)cout<<endl;cin>>cpu>>memory>>nJobs;vector<Job> jobList(nJobs);for(int i=0;i<nJobs;i++){Job job;cin>>job.cpu>>job.memory>>job.arrive>>job.deadline;cin>>job.reward>>job.bonus>>job.punishment;jobList[i]=job;}int income=computeIncome(jobList,cpu,memory,timelineF);cout<<"Case "<<no<<":"<<income<<endl;no++;}}