Printer Queue（poj3125模拟队列）

/*http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27454#problem/A
http://poj.org/problem?id=3125
Printer Queue
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 3337 Accepted: 1813
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


3
1 0
5
4 2
1 2 3 4
6 0
1 1 9 1 1 1
Sample Output


1
2
5
Source


Northwestern Europe 2006
解析：
题意：题意可以理解为有n张牌，每张都有自己的值。当后面有牌的值比当前大时，就把当前牌插到最后去。问当最初第m张拍可以打出来时。总共打出了多少张牌
思路：
模拟队列。将队首元素与当前最大值比较若比其小，则插到队尾，否则打出去。不断更新当前最大值
直至将目标牌打出来。
*/
Accepted 180 KB16 ms C++673 B 2013-07-28 22:31:55

#include<stdio.h>#include<string.h>#include<stack>#include<vector>#include<math.h>#include<queue>#include<algorithm>#include<iostream>using namespace std;const int maxn=100+10;queue<int>q;int a[maxn],b[maxn];int  main(){int T;int n,m,i,j;scanf("%d",&T);while(T--){  while(!q.empty())    {q.pop();    }scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%d",&a[i]);q.push(i);b[i]=a[i];}int t=a[m];sort(b,b+n);int ans=0,max;i=n-1;while(!q.empty()){int x=q.front();q.pop();if(a[x]==b[i]){i--;ans++;if(x==m)break;}elseq.push(x);}printf("%d\n",ans);}return 0;}