【PAT】1017. Queueing at Bank

最新推荐文章于 2025-09-25 09:17:42 发布

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本文通过模拟题探讨了如何使用优先队列优化处理多个窗口的服务时间问题。介绍了两种实现方式，一种直接使用数组记录每个窗口的结束时间并进行更新，另一种则是利用优先队列维护顾客和服务窗口的状态。

考查点：模拟题

思路：用优先队列可提高效率，关键在于维护每个窗口的结束时间数组，每次遍历排队用户时更新结束时间，这里更新时多了个+号结果调了半天

#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#define FOR(i, x, y) for(int i = x; i <= y; i++)
#define rFOR(i, x, y) for(int i = x; i >= y; i--)
#define MAXN 10010
#define oo 0x3f3f3f3f
using namespace std;
struct customer{
    int start,process;
}person[10010];
bool cmp(customer a,customer b){
    return a.start<b.start;
}
int wait[110];

int main()
{
     #ifdef LOCAL
        freopen("data.in","r",stdin);
        freopen("data.out","w",stdout);
    #endif // LOCAL
    int n,k;
     int hh,mm,ss,pp;
    scanf("%d%d",&n,&k);
    int num=0;
    FOR(i,0,n-1)
    {
       scanf("%d:%d:%d %d",&hh,&mm,&ss,&pp);
       int tmp=hh*3600+mm*60+ss;

       if(tmp<=17*3600){
            person[num].process=(pp<60?pp*60:3600);
            person[num].start=tmp;
            num++;

       }
    }
    double sum=0;
    sort(person,person+num,cmp);
   
    FOR(i,0,k-1)
    wait[i]=8*3600;
    FOR(i,0,num-1)
    {
        int wmin=oo;int id;
        FOR(j,0,k-1)
        {
            if(wait[j]<wmin){
                wmin=wait[j];
                id=j;
            }


        }
        if(person[i].start<wait[id]){
                sum+=(wait[id]-person[i].start);
                wait[id]+=person[i].process;

        }else{
            wait[id]=(person[i].start+person[i].process);

        }
        

    }
    if(num==0) printf("0.0");
    else{

    printf("%.1f",sum/60.0/num);
    }

    return 0;
}

优先队列：由于默认最大值。所以要重载运算符

#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#define FOR(i, x, y) for(int i = x; i <= y; i++)
#define rFOR(i, x, y) for(int i = x; i >= y; i--)
#define MAXN 10010
#define oo 0x3f3f3f3f
using namespace std;
struct customer{
    int start,process;
  bool operator<(const customer& a)const{
        if(a.start<start)return true;
        else return false;
  }
}person;
struct window{
    int time;
    bool operator<(const window& a)const{
        if(time>a.time)return true;
        else return false;
    }
}w;
bool cmp(customer a,customer b){
    return a.start<b.start;
}
priority_queue<customer> cu;
priority_queue<window> ww;

int main()
{
     #ifdef LOCAL
        freopen("data.in","r",stdin);
        freopen("data.out","w",stdout);
    #endif // LOCAL
    int n,k;
     int hh,mm,ss,pp;
    scanf("%d%d",&n,&k);
    int num=0;
    FOR(i,0,n-1)
    {
       scanf("%d:%d:%d %d",&hh,&mm,&ss,&pp);
       int tmp=hh*3600+mm*60+ss;

       if(tmp<=17*3600){
            person.process=(pp<60?pp*60:3600);
            person.start=tmp;

            cu.push(person);

       }
    }
    double sum=0;
    num=cu.size();

    FOR(i,0,k-1)
    {
        w.time=8*3600;
        ww.push(w);
    }


    while(!cu.empty())
    {
        w=ww.top();
        int endt=w.time;
        person=cu.top();
        cu.pop();
       if(endt<=person.start){
            endt=person.start+person.process;
            ww.pop();
            w.time=endt;
            ww.push(w);
        }else{
            sum+=(endt-person.start);
            endt+=person.process;
            ww.pop();
            w.time=endt;
            ww.push(w);
        }


    }
    if(num==0) printf("0.0");
    else{

    printf("%.1f",sum/60.0/num);
    }

    return 0;
}