1443. Printer Queue
限制条件
时间限制: 1 秒, 内存限制: 32 兆

题目描述

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.

输入格式

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.

输出格式

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.

样例输入
3
1 0
5
4 2
1 2 3 4
6 0
1 1 9 1 1 1

样例输出
1
2
5

题目的大意是要打印一堆job，每个job有其对应的优先级，当排在最前面的job的优先级不是最高的时候，不打印，把job移动到最后；排在最前面的是最高优先级的时候，打印。多个测试用例，每个用例输入包含n m，代表job个数，和你要打印的job在原始队列中的位置，接下来是一串数字，代表对应位置的job的优先级。你需要figure out指导你的job打印完成一共花了多长时间。每一次打印花费1时间，移动不耗时。

这个问题直接用队列模拟就行啦，下面是我的代码:

//1443. Printer Queue
// 2016.10.13

#include <iostream> 
#include <queue>
#include <algorithm>
using namespace std;

queue<int> q1,q2;   // q1 记录各个位置，q2 记录按优先级排好的值 
int arr[104],sarr[104];

bool cmp(const int &a,const int &b)   //比较函数 
	if(a>b)
		return 1;
	else
		return 0;
}
int main()
{
	int test;
	cin>>test;
	
	int n,m;
	for(int i=1;i<=test;i++)
	{
		cin>>n>>m;
		for(int i=0;i<n;i++)
		{
			cin>>arr[i];
			sarr[i]=arr[i];
			q1.push(i);
		}
		
		sort(sarr,sarr+n,cmp);
		for(int i=0;i<n;i++)
		{
			q2.push(sarr[i]);
		}
	
		int sum=0;
		
		while(arr[q1.front()] != q2.front() || q1.front() != m)   
		{
			
			if(arr[q1.front()] != q2.front())  //队列头 不为最高级的时候 
			{
				
				q1.push(q1.front()) ;
				q1.pop();
			}
			if(arr[q1.front()] == q2.front()&& q1.front() != m)   // 队列头为最高级，但不是我们要求的值 
			{
				
				q1.pop();
				q2.pop();
				sum++;
				
			}
			
		}
		while(!q1.empty())   // 清栈 
			q1.pop();
		while(!q2.empty())
			q2.pop();
		cout<<sum+1<<endl;
	}
	return 0;
}