算法学习【13】—— 1443. Printer Queue
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题目来源:http://soj.me/1443
1443. Printer Queue
Description
The only printer in the computer science students' union is experiencing an extremely heavy workload. Sometimes there are a hundred jobs in the printer queue and you may have to wait for hours to get a single page of output.
Because some jobs are more important than others, the Hacker General has invented and implemented a simple priority system for the print job queue. Now, each job is assigned a priority between 1 and 9 (with 9 being the highest priority,
and 1 being the lowest), and the printer operates as follows.
- The first job J in queue is taken from the queue.
- If there is some job in the queue with a higher priority than job J, thenmove J to the end of the queue without printing it.
- Otherwise, print job J (and do not put it back in the queue).
Your problem with the new policy is that it has become quite tricky to determine when your print job will actually be completed. You decide to write a program to figure this out. The program will be given the current queue (as a list of priorities) as well as the position of your job in the queue, and must then calculate how long it will take until your job is printed, assuming that no additional jobs will be added to the queue. To simplifymatters, we assume that printing a job always takes exactly one minute, and that adding and removing jobs from the queue is instantaneous.
Input
One line with a positive integer: the number of test cases (at most 100). Then for each test case:
- One line with two integers n and m, where n is the number of jobs in the queue (1 ≤ n ≤ 100) and m is the position of your job (0 ≤ m ≤ n −1). The first position in the queue is number 0, the second is number 1, and so on.
- One linewith n integers in the range 1 to 9, giving the priorities of the jobs in the queue. The first integer gives the priority of the first job, the second integer the priority of the second job, and so on.
Output
For each test case, print one line with a single integer; the number of minutes until your job is completely printed, assuming that no additional print jobs will arrive.
Sample Input
31 054 21 2 3 46 01 1 9 1 1 1
Sample Output
125
思路:直接模拟实现,一个计数数组用于存放各个权值的作业个数,然后另一个数组用于存放作业。不断遍历扫描……
代码:
#include <iostream>using namespace std;int value[100];int counter[10];int myjob;int jobnumber;int main(){int cast;int i;int maxpriority;int scaner;int result;cin >> cast;while (cast--){cin >> jobnumber >> myjob;result = 0;scaner = 0;for (i = 1; i < 10; i++){counter[i] = 0;}for (i = 0; i < jobnumber; i++){value[i] = -1;}for (i = 0; i < jobnumber; i++){cin >> value[i];counter[value[i]]++;}for (i = 9; i > 0; i--){if (counter[i] != 0){maxpriority = i;break;}}while (maxpriority != value[myjob] || myjob != scaner){if (maxpriority == value[scaner]){result++;value[scaner] = -1;counter[maxpriority]--;while (0 == counter[maxpriority]){maxpriority--;}}scaner = (scaner + 1) % jobnumber;while (-1 == value[scaner]){scaner = (scaner + 1) % jobnumber;}}cout << result + 1<< endl;}return 0;}
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