【暑假】[实用数据结构]UVAlive 3135 Argus

UVAlive 3135 Argus

Argus

Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

Submit Status

Description

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logs and telephone call records. Likewise, queries over streams run continuously over a period of time and incrementally return new results as new data arrives. For example, a temperature detection system of a factory warehouse may run queries like the following.

Query-1: ?Every five minutes, retrieve the maximum temperature over the past five minutes.? Query-2: ?Return the average temperature measured on each floor over the past 10 minutes.?

We have developed a Data Stream Management System called Argus, which processes the queries over the data streams. Users can register queries to the Argus. Argus will keep the queries running over the changing data and return the results to the corresponding user with the desired frequency.

For the Argus, we use the following instruction to register a query:

Register Q_num Period

Q_num (0 < Q_num ≤ 3000) is query ID-number, and Period (0 < Period ≤ 3000) is the interval between two consecutive returns of the result. After Period seconds of register, the result will be returned for the first time, and after that, the result will be returned every Period seconds.

Here we have several different queries registered in Argus at once. It is confirmed that all the queries have different Q_num. Your task is to tell the first K queries to return the results. If two or more queries are to return the results at the same time, they will return the results one by one in the ascending order of Q_num.

Input

The first part of the input are the register instructions to Argus, one instruction per line. You can assume the number of the instructions will not exceed 1000, and all these instructions are executed at the same time. This part is ended with a line of ?#?.

The second part is your task. This part contains only one line, which is one positive integer K (≤ 10000).

Output

You should output the Q_num of the first K queries to return the results, one number per line.

Sample Input

Register 2004 200
Register 2005 300
#
5

Sample Output

2004
2005
2004
2004
2005

--------------------------------------------------------------------------------------------------------------------------------------------------------

思路:  优先队列模拟。时间小的优先级高,每次选取time最小的出队记录Qnum后修改time重新入队(意味着该任务进入下一轮)  注意:优先队列中重载运算符比较特殊,我这样理解:STL中默认priority_queue为大根堆,而我们所需要的是小根堆,因此将 < 重载为相反符号。

代码:

 1 //处理器模拟
 2 #include<cstdio>
 3 #include<queue>
 4 #define FOR(a,b,c) for(int a=(b);a<(c);a++)
 5 using namespace std;
 6
 7 struct Node{
 8     int qnum,period,time;
 9     bool operator < (const Node& rhs) const{
10      return time>rhs.time || (time==rhs.time && qnum>rhs.qnum);
11     }
12 };
13 //优先队列默认为大根堆,而需要的是“小根堆 ”因此相反地定义 <
14
15 int main(){
16 priority_queue<Node> Q;
17 char s[20];
18 Node x;
19
20   while(scanf("%s",s) && s[0] != ‘#‘){
21       scanf("%d%d",&x.qnum,&x.period);
22       x.time=x.period;   //time_init
23     Q.push(x);
24   }
25   int k; scanf("%d",&k);
26   while(k--){
27      x=Q.top(); Q.pop();
28      printf("%d\n",x.qnum);
29      x.time += x.period;    //更改为下一轮的时间重新放回优先队列
30      Q.push(x);
31   }
32   return 0;
33 }
				
时间: 2024-10-10 09:16:48

【暑假】[实用数据结构]UVAlive 3135 Argus的相关文章

UVALive - 3135 - Argus (优先队列!!)

UVALive - 3135 Argus Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financi

【优先队列之多路归并】UVALive 3135 Argus

UVALive 3135 Argus http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18684 题意 编写一个系统执行一系列Register命令:Register Q_num Period,每个命令执行周期是Period,执行事件Q_num:如果事件同时发生,优先执行Q_num小的. 示例 Sample Input Register 2004 200 Register 2005 300 # 5 Sample Output

uva11997 K Smallest Sums&amp;&amp;UVALive 3135 Argus(优先队列,多路归并)

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<stack> #in

【暑假】[实用数据结构]UVAlive 3026 Period

UVAlive 3026 Period 题目: Period Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive),

【暑假】[实用数据结构]UVAlive 3942 Remember the Word

UVAlive 3942 Remember the Word 题目: Remember the Word Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Neal is very curious about combinatorial problems, and now here comes a problem about words. Know

【暑假】[实用数据结构]UVAlive 3027 Corporative Network

UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent

【暑假】[实用数据结构]UVAlive 3644 X-Plosives

UVAlive X-Plosives 思路:    “如果车上存在k个简单化合物,正好包含k种元素,那么他们将组成一个易爆的混合物”  如果将(a,b)看作一条边那么题意就是不能出现环,很容易联想到Kruskal算法中并查集的判环功能(新加入的边必须属于不同的两个集合否则出现环),因此本题可以用并查集实现.模拟装车过程即可. 代码: 1 #include<cstdio> 2 #include<cstring> 3 #define FOR(a,b,c) for(int a=(b);a

【暑假】[实用数据结构]UVAlive

题目:   Dominating Patterns Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description The archaeologists are going to decipher a very mysterious ``language". Now, they know many language patterns; each patt

UVALIVE 3135 Argus

#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #inc