UVA 1203 - Argus 题目链接 题意:给定一些注冊命令.表示每隔时间t,运行一次编号num的指令.注冊命令结束后.给定k.输出前k个运行顺序 思路:用优先队列去搞,任务时间作为优先级.每次一个任务出队后,在把它下次运行作为一个新任务入队就可以 代码: #include <cstdio> #include <cstring> #include <queue> using namespace std; char str[10]; struct Task { in…

option=com_onlinejudge&Itemid=8&page=show_problem&problem=3644" target="_blank" style="">题目链接:uva 1203 - Argus 题目大意:一个成为Argus的系统.支持一个Register命令 Register Q_num Period 该命令为一个祖冊出发器,每Period产生一次Q_num的时间.要求模拟k个事件,假设多个事件同一…

简单的优先队列的应用: 代码: #include<queue> #include<cstdio> using namespace std; struct node { int num; int ti; int period; bool operator<(const node &t)const { if(ti==t.ti)return num>t.num; return ti>t.ti; } }; priority_queue< node >q…

题目链接: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3148 题意:给K个数组,每个数组含有K个整数,从每个数组中各选一个数加起来,得到一个sum,这样的选法一共有K^k种,现在求这样的和中最小的K个sum. 解法:先考虑两个数组的情形 假设A1<=A2<=`````<=AK B1<=B2<=`````&…

3135 - Argus A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensordata, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logsand telephone call records. Likewise…

直接用一个优先队列去模拟Huffman树的建立过程. 每次取优先队列前两个数,然后累加其和,把这个和在放入到优先队列中去. #include <cstdio> #include <queue> using namespace std; int main() { int n; && n) { priority_queue<int, vector<int>, greater<int> > q; ; ; i < n; i++) {…

题目链接: http://poj.org/problem?id=2051 思路分析: 优先级问题,使用优先队列求解:当执行某个任务后,再增加一个任务到队列中, 该任务的优先级为执行任务的时间加上其时间间隔,如此反复直到求出前K个执行任务. 代码: #include <iostream> #include <queue> using namespace std; struct Argu { int QNum; int period; int time; bool operator&l…

Problem Description Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ...shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500’th ugly…

#include <iostream> #include <string> #include <queue> using namespace std; struct In { int id; int ReturnTime; int period; In(int i, int p):id(i), ReturnTime(p), period(p){} //重载写在在里面又怎么写? }; bool operator < (const In& x, const I…

解法和合并果子是一样的, 每次取最小的两个, 更新答案, 加入队列 #include<cstdio> #include<queue> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; int main() { int n, x; while(~scanf("%d", &n) && n) { priority_queue<int, v…