POJ 3125 Printer Queue 数据结构 队列
2015-08-18 18:49
633 查看
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).
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
Sample Output
Source
Northwestern Europe 2006
inline快const快 tle------>0ms
ACcode
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4329 | Accepted: 2269 |
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
inline快const快 tle------>0ms
ACcode
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=102; inline int aabs(int a){ return a<0?-a:a; } int a[maxn],st,ed; int main(){ int loop; scanf("%d",&loop); while(loop--){ int num,pos; scanf("%d %d",&num,&pos); for(int i=0;i<num;++i)scanf("%d",&a[i]); a[pos]=-a[pos]; st=0; ed=num; int cns=0; while((ed+1)%maxn!=st){ int k=a[st]; st=(st+1)%maxn; bool flag=true; for(int i=st;i!=ed;i=(i+1)%maxn) if(aabs(a[i])>aabs(k)){ flag=false; a[ed]=k; ed=(ed+1)%maxn; break; } if(flag){ cns++; if(k<0){ printf("%d\n",cns); break; } } } } return 0; }
相关文章推荐
- 2015-8-18数据结构-动态规划-矩阵乘法次数最少
- 数据结构—排序总结
- 数据结构实验之二叉树的建立与遍历
- POJ 2259 Team Queue 数据结构 队列
- 浅谈数据结构-树
- 数据结构实验之求二叉树后序遍历和层次遍历
- 传说中的数据结构 SDUT 2556
- 数据结构上机测试4.1:二叉树的遍历与应用1--知先序和中序求后序
- c语言实现数据结构中的链式表
- c语言实现数据结构中的顺序表
- c语言实现数据结构中的哈希表
- c语言实现数据结构中的队列
- c语言实现数据结构中的栈
- 数据结构—进制间转换
- 浅谈数据结构-Boyer-Moore算法
- 数据结构五:二叉树的递归遍历,二叉树的叶子节点个数,二叉树的拷贝操作基础学习
- 浅谈数据结构-字符串匹配
- 数据结构——全面学习哈希
- 2015-8-18数据结构-分冶-整数划分问题
- Codeforces Round #286 (Div. 1) B. Mr. Kitayuta's Technology (强连通分量)