u013172314 于 2015.05.25 14:45 提问

PAT 1017. Queueing at Bank

【题目链接】

``````#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;

int n, k;
int leaveTime[10005];
int waitTime[10005];
int window[105]; //存放对应窗口当前服务的客户编号

struct Customer
{
int arriveTime;
int processTime;
}buf[10005];

bool cmp(Customer a, Customer b)
{
return a.arriveTime < b.arriveTime;
}

int main()
{
//freopen("in_1017.txt", "r", stdin);
int hour, minute, second;
cin >> n >> k;
for(int i = 0; i < n; i++)
{
cin >> hour;
getchar();
cin >> minute;
getchar();
cin >> second >> buf[i].processTime;
buf[i].processTime *= 60;
buf[i].arriveTime = hour * 3600 + minute * 60 + second;
}
sort(buf, buf + n, cmp);
int cus;
int num = 0; //在17点之后到达的顾客数量
for(int i = 0; i < k && i < n; i++)
{
cus = i;
window[i] = i;
if(buf[i].arriveTime < 8 * 3600)
{
leaveTime[i] = 8 * 3600 +  buf[i].processTime;
waitTime[i] = 8 * 3600 - buf[i].arriveTime;
}
else if(buf[i].arriveTime <= 17 * 3600)
{
leaveTime[i] = buf[i].arriveTime +  buf[i].processTime;
waitTime[i] = 0;
}
else
{
waitTime[i] = -1;
num++;
continue;
}
}
int min_leave; //最早的离开时间********************
int min_win; //每次离开的客户对应的窗口序号
for(int i = cus + 1; i < n; i++)
{
if(buf[i].arriveTime > 17 * 3600)
{
waitTime[i] = -1;
num++;
continue;
}
min_leave = 24 * 3600 + 100;
for(int j = 0; j < k; j++)
{
int u = window[j]; //u保存当前窗口服务的客户编号
if(leaveTime[u] < min_leave)
{
min_leave = leaveTime[u];
cus = u;
min_win = j;
}
}
window[min_win] = i;
//当有窗口空闲, 但是下一位顾客还未到来时, 窗口需要等待;
if(leaveTime[cus] < buf[i].arriveTime) leaveTime[cus] = buf[i].arriveTime;
waitTime[i] = leaveTime[cus] - buf[i].arriveTime;
leaveTime[i] = leaveTime[cus] +  buf[i].processTime;
}
double average = 0;
for(int i = 0; i < n; i++)
if(waitTime[i] >= 0) average += waitTime[i];
else break;
if(n == num) num = 0; //若全部顾客在17点之后到达,下面语句分母为0
printf("%.1lf\n", average / 60 / (n - num));
}
``````