[sicily]1443. Printer Queue

1443. Printer Queue

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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).

In this way, all those importantmuffin recipes that the Hacker General is printing get printed very quickly. Of course, those annoying term papers that others are printing may have to wait for quite some time to get printed, but that's life. 

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

简单队列应用题。题目大意是，要求每次队列中最大的优先级的元素出队，判断如果队首不是最大优先级元素，则用放到队尾，如果是最大优先级否则输出。给出初始队列指定位置元素，问输出该元素的时间。直接模拟操作可以完成。注意，两个小技巧：1，利用struck 保存初始队列元素相对位置信息和优先级信息； 2，设置一个数组保存初始队列优先级，用于比较，队列中已经不存在的则优先级为0. 具体代码如下：

#include <iostream>#include <queue>using namespace std;struct node{    int position;<span style="white-space:pre"></span>//初始位置    int priority;<span style="white-space:pre"></span>//优先级大小}; int main(){    int t;    cin>>t;    while(t--)    {        queue<node> jobs;        int job[101];// 保存初始队列相对位置的各个元素优先级        int n,m;        cin>>n>>m;        struct node tmp;                for(int i=0; i<n; i++) //输入队列保存</span>        {                        tmp.position = i;            cin>>job[i];            tmp.priority = job[i];            jobs.push(tmp);        }        int count = 0;  //计数器        while(1)        {            bool flag = 0;            tmp = jobs.front();             jobs.pop();//取队首元素，并出队，下面判断这个元素是否需要队尾入队。            for(int i=0; i<n; i++)<span style="white-space:pre"></span>            {                if(job[i] > tmp.priority)                {                    flag = 1;//存在比当前队首元素优先级更高的，所以设置flag = 1，表示需要入队。                    break;                }               }            if(flag)            {                jobs.push(tmp);            }            else//flag == 0，表示当前队首元素优先级最大，出队后无需入队            {                count++;//输出 计数器 加 1                 job[tmp.position] = 0;// 初始优先级数组优先级设为 0，表示失效；                if(tmp.position == m)                {                    cout<<count<<endl;                    break;                }               }        }    }     //system("pause");    return 0;   }