PAT 1014 Waiting in Line (30)
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#include <iostream>#include <cstring>#include <queue>#define inf 10000using namespace std;typedef struct window{int endTime;queue<int> q;}window;typedef struct customer{int startTime;int endTime;int wasteTime;}customer;window win[22];//未初始化结构体成员变量都是0 customer cus[1002];int main(){int N,M,K,Q,i,j,minTime,location;//N个窗口,每个窗口最多M人排队,现在来了K个人员 int wasteTime[1002],queryNum[1002];cin>>N>>M>>K>>Q;for(i=1;i<=K;i++)cin>>cus[i].wasteTime;for(i=0;i<Q;i++)cin>>queryNum[i];/*下面尽量选用下标从1开始,因为涉及到顾客的编号,没有编号0,最小编号从1开始的*/for(i=1,j=1;i<=N*M&&i<=K;++i,j=i%N){//初始化一开始黄线以内等候排队服务的人,第i个客户等候在第j个窗口 cus[i].startTime=win[j].endTime;//一开始时候win[j].endTime都是0 cus[i].endTime=cus[i].startTime+cus[i].wasteTime;win[j].endTime=cus[i].wasteTime+win[j].endTime;win[j].q.push(i);//第i个人进入第j个窗口 }for(;i<=K;i++){//前面已经安排完毕N*M个人,接着安排minTime=inf;for(j=0;j<N;j++){int lastNum=win[j].q.front();//现在要从这N只队伍当中找出服务最先结束的那一只 if(cus[lastNum].endTime<minTime)//找出每个队伍当中最后结束花费最短时间最短那个 {minTime=cus[lastNum].endTime;location=j;//把这个人安置在第j个窗口 }}cus[i].startTime=win[location].endTime;//又进来一个客户,重新更新上面数组 cus[i].endTime=cus[i].startTime+cus[i].wasteTime;win[location].endTime=cus[i].wasteTime+win[location].endTime;win[location].q.pop();//上面一个人服务结束 win[location].q.push(i); }int hour,minute;for(i=0;i<Q;i++)if(cus[queryNum[i]].startTime+8*60>=17*60)cout<<"Sorry"<<endl;else{hour=cus[queryNum[i]].endTime/60+8;minute=cus[queryNum[i]].endTime%60;if(hour<10)cout<<"0";cout<<hour<<":";if(minute<10)cout<<"0";cout<<minute<<endl;} return 0;}
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