九度OJ 1326:Waiting in Line(排队) (模拟)
时间限制:1 秒
内存限制:32 兆
特殊判题:否
提交:220
解决:64
- 题目描述:
-
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:
- The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line
behind the yellow line. - Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
- Customer[i] will take T[i] minutes to have his/her transaction processed.
- The first N customers are assumed to be served at 8:00am.
Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.
For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer1 is
served at window1 while customer2 is served at window2. Customer3 will
wait in front of window1 and customer4 will wait in front of window2. Customer5 will
wait behind the yellow line.At 08:01, customer1 is done and customer5 enters the line in front of window1 since that line seems shorter now. Customer2 will
leave at 08:02, customer4 at 08:06, customer3 at 08:07, and finally customer5 at 08:10. - The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line
- 输入:
-
Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (<=20, number of windows), M (<=10, the maximum capacity of each line inside the yellow line), K (<=1000, number of customers), and Q (<=1000, number of customer
queries).The next line contains K positive integers, which are the processing time of the K customers.
The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.
- 输出:
-
For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served
before 17:00, you must output "Sorry" instead.
- 样例输入:
-
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
- 样例输出:
-
08:07
08:06
08:10
17:00
Sorry
思路:
模拟题,排队题当然数据结构用队列比较好。
用C语言太苦逼了,还要自己写一个队列,还是C++比较好,以后要用C++了。
需要注意的是:即使是17点之前开始办业务,17点之后仍然没有办完就是sorry。
代码:
#include <stdio.h>
#include <stdlib.h> #define N 21
#define M 11
#define K 1001 typedef struct node1 {
int proc;
int begin;
} Cust; typedef struct node2 {
Cust *cust[M];
int front;
int rear;
} Queue; int isEmpty(Queue *q)
{
return (q->front == q->rear);
} int isFull(Queue *q)
{
return (q->front == (q->rear+1)%M);
} int countQueue(Queue *q)
{
return (q->rear + M - q->front) % M;
} void push(Queue *q, Cust *val)
{
q->cust[q->rear] = val;
q->rear = (q->rear+1)%M;
} Cust *pop(Queue *q)
{
int front = q->front;
q->front = (q->front+1)%M;
return q->cust[front];
} Cust *top(Queue *q)
{
return q->cust[q->front];
} int main(void)
{
int n, m, k, q, i, j;
Queue *queue[N];
Cust *cust[K];
int query[K];
int maxTime = (17-8)*60; while (scanf("%d%d%d%d", &n, &m, &k, &q) != EOF)
{
for (i=0; i<n; i++)
{
queue[i] = (Queue *)malloc(sizeof(Queue));
queue[i]->front = queue[i]->rear = 0;
}
for (i=0; i<k; i++)
{
cust[i] = (Cust *)malloc(sizeof(Cust));
scanf("%d", &cust[i]->proc);
cust[i]->begin = maxTime+1;
}
for (i=0; i<q; i++)
scanf("%d", &query[i]); // init state
int inQueue = 0;
int outQueue = k;
for (i=0; i<m; i++)
{
for (j=0; j<n; j++)
{
inQueue = i*n + j;
if (inQueue >= k)
break;
if (i == 0)
cust[inQueue]->begin = 0;
push(queue[j], cust[inQueue]);
}
if (inQueue >= k)
break;
}
inQueue = (m*n <= k) ? (inQueue+1) : inQueue;
outQueue = k - inQueue;
//printf("inQueue=%d, outQueue=%d, k=%d\n", inQueue, outQueue, k); // simulate the whole process
int time;
for (time=0; time<=maxTime; time++)
{
for (i=0; i<n; i++)
{
if (isEmpty(queue[i]))
break;
Cust *c = top(queue[i]);
if (c->begin + c->proc == time)
{
pop(queue[i]);
if (outQueue > 0)
{
push(queue[i], cust[k-outQueue]);
outQueue --;
}
if (!isEmpty(queue[i]))
{
top(queue[i])->begin = time;
//printf("i=%d, time=%d\n", i, time);
}
}
}
if (isEmpty(queue[0]))
break;
} for (i=0; i<q; i++)
{
int id = query[i] - 1;
int endTime = cust[id]->begin + cust[id]->proc;
if (endTime > maxTime)
printf("Sorry\n");
else
printf("%02d:%02d\n", 8 + endTime/60, endTime%60);
}
} return 0;
}
/**************************************************************
Problem: 1326
User: liangrx06
Language: C
Result: Accepted
Time:10 ms
Memory:916 kb
****************************************************************/
九度OJ 1326:Waiting in Line(排队) (模拟)的更多相关文章
- 九度OJ 1110:小白鼠排队 (排序)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:1734 解决:1054 题目描述: N只小白鼠(1 <= N <= 100),每只鼠头上戴着一顶有颜色的帽子.现在称出每只白鼠的 ...
- 九度OJ 1068 球半径和数量 (模拟)
题目1068:球的半径和体积 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:4797 解决:1696 题目描写叙述: 输入球的中心点和球上某一点的坐标,计算球的半径和体积 输入: 球的中心 ...
- 九度OJ 1067 n的阶乘 (模拟)
题目1067:n的阶乘 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:5666 解决:2141 题目描写叙述: 输入一个整数n,输出n的阶乘 输入: 一个整数n(1<=n<=2 ...
- 九度OJ 1183 守形数 (模拟)
题目1183:守形数 时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:2663 解决:1424 题目描写叙述: 守形数是这样一种整数.它的平方的低位部分等于它本身. 比方25的平方是625. ...
- 九度OJ 1355:扑克牌顺子 (模拟)
时间限制:2 秒 内存限制:32 兆 特殊判题:否 提交:1676 解决:484 题目描述: LL今天心情特别好,因为他去买了一副扑克牌,发现里面居然有2个大王,2个小王(一副牌原本是54张^_^). ...
- 九度OJ 1334:占座位 (模拟)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:864 解决:202 题目描述: sun所在学校的教室座位每天都是可以预占的. 一个人可以去占多个座位,而且一定是要连续的座位,如果占不到他所 ...
- 【九度OJ】题目1124:Digital Roots 解题报告
[九度OJ]题目1124:Digital Roots 解题报告 标签(空格分隔): 九度OJ 原题地址:http://ac.jobdu.com/problem.php?pid=1124 题目描述: T ...
- 【九度OJ】题目1433:FatMouse 解题报告
[九度OJ]题目1433:FatMouse 解题报告 标签(空格分隔): 九度OJ http://ac.jobdu.com/problem.php?pid=1433 题目描述: FatMouse pr ...
- 【九度OJ】题目1439:Least Common Multiple 解题报告
[九度OJ]题目1439:Least Common Multiple 解题报告 标签(空格分隔): 九度OJ 原题地址:http://ac.jobdu.com/problem.php?pid=1439 ...
随机推荐
- EM算法和GMM模型推导
- oracle 12C SYS,SYSTEM用户的密码都忘记或是丢失
密码 conn / as sysdba alter user system identified by Abcd1234; manual script first -->manual_scrip ...
- Eclipse web项目导入Intellij 并且部署
一.导入自己的web项目 步骤:File->New->Project from Existing Source... 二.选择项目的所在位置,点击"OK";接着如下图所 ...
- 微信小程序 - 如何通过button按钮实现分享(转发)功能
小程序官方API https://developers.weixin.qq.com/miniprogram/dev/framework/app-service/page.html#%E9%A1%B5% ...
- sql NextResult()多个结果集
转自 http://blog.csdn.net/limlimlim/article/details/8626898 注意:当SQL语句中出现两条Select语句,例如:string sql = &q ...
- apache环境下禁止某文件夹内运行PHP脚本、禁止访问文件或目录执行权限的设置方法
apache环境下禁止某文件夹内运行PHP脚本.禁止访问文件或目录执行权限的设置方法 首先我们来看两段对上传目录设置无权限的列子,配置如下: <Directory "要去掉PHP执 ...
- debounce 防抖动函数
http://lodash.think2011.net/debounce _.debounce(func, [wait=0], [options]) 创建一个防抖动函数. 该函数会在 wait 毫秒后 ...
- Activity的启动流程分析
Activity是Android应用程序的四大组件之中的一个,负责管理Android应用程序的用户界面,一般一个应用程序中包括非常多个Activity,他们可能执行在一个进程中.也可能执行在不同的进程 ...
- R 包的安装,使用,更新
R包的使用方法 包就是提供了种类繁多的函数,当然还有它的一些数据集,可以使用这些函数来操作这些数据集,来学习使用. library(),当前的工作环境里,可以使用的包 包的帮助文档:help(pack ...
- (六)Thymeleaf的 th:* 属性之—— th: ->text& utext& href
th:*使用原因: for the sake of simplicity and compactness of the code samples(简化代码) the th:*notation is m ...