PAT (Advanced) 1017. Queueing at Bank (25)
2017-09-21 23:18
1016 查看
原题:1017. Queueing at Bank (25)
解题思路:
1.将所有顾客的时间转化为相对00:00 : 00 的秒, 然后进行排序。
2.每次寻找最早结束的窗口分配给当前顾客,计算等待时间。
3.按照顾客到达时间进行动态更新服务窗口的结束时间,注意过滤17:00:00以后来的顾客。
注:当没有顾客时直接输出0.0; 寻找最早服务结束的窗口时,注意把初始的min值设大一些,一开始我设到晚上9点,结果最后一个测试点没过; 只要17:00 前到的顾客,不管窗口服务有没有在17:00前结束,都要为其服务。
C++代码如下:
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 10010;
struct Customer
{
int serveTime;
int arriveTime;
} customer[maxn];
int win[101];
int getTime(int h, int m, int s)
{
return h * 3600 + m * 60 + s;
}
bool cmp(Customer a, Customer b)
{
return a.arriveTime < b.arriveTime;
}
int main()
{
int n, k;
while(scanf("%d%d", &n, &k) != EOF)
{
int st = getTime(8, 0, 0);
int ed = getTime(17, 0, 0);
for(int i = 0; i < n; i++)
{
int h, m, s, serve;
scanf("%d:%d:%d %d", &h, &m, &s, &serve);
customer[i].serveTime = serve * 60;
customer[i].arriveTime = getTime(h, m, s);
}
//初始化窗口
for(int i = 0; i < k; i++)
{
win[i] = st;
}
sort(customer, customer + n, cmp);
int cnt = 0, total = 0;
for(int i = 0; i < n; i++)
{
//对到达顾客进行预处理,超时服务时间转为一小时
if(customer[i].arriveTime > ed)
break;
if(customer[i].serveTime > 3600)
customer[i].serveTime = 3600;
//寻找服务窗口
int u = -1, MIN = 100000000;
for(int j = 0; j < k; j++)
{
if(win[j] < MIN)
{
u = j;
MIN = win[j];
}
}
if(customer[i].arriveTime < win[u])
{
total += (win[u] - customer[i].arriveTime);
win[u] = win[u] + customer[i].serveTime;
}
else
{
win[u] = customer[i].arriveTime + customer[i].serveTime;
}
cnt++;
}
if(cnt == 0)
printf("0.0\n");
else
printf("%.1f\n", double(total) / 60 / cnt);
}
return 0;
}
解题思路:
1.将所有顾客的时间转化为相对00:00 : 00 的秒, 然后进行排序。
2.每次寻找最早结束的窗口分配给当前顾客,计算等待时间。
3.按照顾客到达时间进行动态更新服务窗口的结束时间,注意过滤17:00:00以后来的顾客。
注:当没有顾客时直接输出0.0; 寻找最早服务结束的窗口时,注意把初始的min值设大一些,一开始我设到晚上9点,结果最后一个测试点没过; 只要17:00 前到的顾客,不管窗口服务有没有在17:00前结束,都要为其服务。
C++代码如下:
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 10010;
struct Customer
{
int serveTime;
int arriveTime;
} customer[maxn];
int win[101];
int getTime(int h, int m, int s)
{
return h * 3600 + m * 60 + s;
}
bool cmp(Customer a, Customer b)
{
return a.arriveTime < b.arriveTime;
}
int main()
{
int n, k;
while(scanf("%d%d", &n, &k) != EOF)
{
int st = getTime(8, 0, 0);
int ed = getTime(17, 0, 0);
for(int i = 0; i < n; i++)
{
int h, m, s, serve;
scanf("%d:%d:%d %d", &h, &m, &s, &serve);
customer[i].serveTime = serve * 60;
customer[i].arriveTime = getTime(h, m, s);
}
//初始化窗口
for(int i = 0; i < k; i++)
{
win[i] = st;
}
sort(customer, customer + n, cmp);
int cnt = 0, total = 0;
for(int i = 0; i < n; i++)
{
//对到达顾客进行预处理,超时服务时间转为一小时
if(customer[i].arriveTime > ed)
break;
if(customer[i].serveTime > 3600)
customer[i].serveTime = 3600;
//寻找服务窗口
int u = -1, MIN = 100000000;
for(int j = 0; j < k; j++)
{
if(win[j] < MIN)
{
u = j;
MIN = win[j];
}
}
if(customer[i].arriveTime < win[u])
{
total += (win[u] - customer[i].arriveTime);
win[u] = win[u] + customer[i].serveTime;
}
else
{
win[u] = customer[i].arriveTime + customer[i].serveTime;
}
cnt++;
}
if(cnt == 0)
printf("0.0\n");
else
printf("%.1f\n", double(total) / 60 / cnt);
}
return 0;
}
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