http://poj.org/problem?id=2051
Argus
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 9716 | Accepted: 4543 |
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
Source
水题拉，就是学习下优先队列结构体咋写，注意重载时候大于其实是小的优先，因为stl默认是大根堆。。
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-8;
const double pi=acos(-1.0);
const int INF=0x7fffffff;
const LL inf=(((LL)1)<<61)+5;
struct item{
int qnum,period,time;
bool operator < (const item& a) const{
return time>a.time||(time==a.time&&qnum>a.qnum); //这里注意大于号实际是time小的优先
}
};
char s[23];
int main()
{
priority_queue<item>q;
while(~scanf("%s",s)&&strcmp(s,"#")!=0)
{
item cur;
scanf("%d%d",&cur.qnum,&cur.period);
cur.time=cur.period;
q.push(cur);
}
int k;
scanf("%d",&k);
while(k--)
{
item cur=q.top();
q.pop();
printf("%d\n",cur.qnum);
cur.time+=cur.period;
q.push(cur);
}
return 0;
}