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2013
11-10

POJ 2051 Argus [解题报告] Java

Argus

问题描述 :

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.

输入:

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).

输出:

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

样例输入:

Register 2004 200
Register 2005 300
#
5

样例输出:

2004
2005
2004
2004
2005

解题代码:

import java.util.Scanner;

public class Main {
 private static Query[] query = new Query[1002];

 public static void main(String[] args) {
  Scanner sc = new Scanner(System.in);

  int i = 0;
  while (sc.hasNext()) {
   String s = sc.nextLine();
   if (s.equals("#")) {
    int k = sc.nextInt();
    slove(k, i);
    break;
   }
   String[] str = s.split(" ");
   Query q = new Query();
   q.d = Integer.parseInt(str[1].trim());
   q.w = Integer.parseInt(str[2].trim());
   q.p = q.w;
   insertHeap(q, i);
   i++;
  }
 }

 public static void slove(int k, int n) {
  // System.out.println("start.........");
 
  while (k > 0) {
   System.out.println(query[0].d);
   query[0].w += query[0].p;
   // System.out.println(query[0].w);
   keapHeap(n);
   --k;
  }

 }

 public static void keapHeap(int n) {

  Query q = query[0];
  // System.out.println(q.w + ";" + q.d);
  int i = 0;
  int j = (i << 1) + 1;
  while (j <= n - 1) {
   if (j < n - 1 && query[j].w > query[j + 1].w)
    j++;
   else if (j < n - 1 && query[j].w == query[j + 1].w
     && query[j].d > query[j + 1].d) {
    j++;
   }
   if (query[j].w < q.w) {
    query[i] = query[j];
    i = j;
   } else if (query[j].w > q.w)
    break;
   else if (query[j].w == q.w && query[j].d < q.d) {
    query[i] = query[j];
    i = j;
   } else
    break;
   j = (i << 1) + 1;
  }
  query[i] = q;
 }

 public static void insertHeap(Query q, int n) {
  // System.out.println("n = " + n);
  if (n == 0) {
   query[0] = q;
  } else {

   int i = n;
   int j = (i - 1) >> 1;

   while (j >= 0) {
    if (query[j].w > q.w) {
     query[i] = query[j];
     i = j;
    } else if (query[j].w == q.w && query[j].d > q.d) {
     query[i] = query[j];
     i = j;
    } else
     break;
    j = (i - 1) >> 1;
   }
   query[i] = q;
  }
 }

 static class Query {
  private int w;
  private int p;
  private int d;
 }
}

  1. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”