概述
Description
Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example.
In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue.
Your task is to write a program that simulates such a team queue.
In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue.
Your task is to write a program that simulates such a team queue.
Input
The input will contain one or more test cases. Each test case begins with the number of teams t (1<=t<=1000). Then t team descriptions follow, each one consisting of the number of elements belonging to the team and the elements themselves. Elements are integers in the range 0 - 999999. A team may consist of up to 1000 elements.
Finally, a list of commands follows. There are three different kinds of commands:
The input will be terminated by a value of 0 for t.
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.
Finally, a list of commands follows. There are three different kinds of commands:
- ENQUEUE x - enter element x into the team queue
- DEQUEUE - process the first element and remove it from the queue
- STOP - end of test case
The input will be terminated by a value of 0 for t.
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.
Output
For each test case, first print a line saying "Scenario #k", where k is the number of the test case. Then, for each DEQUEUE command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one.
Sample Input
2 3 101 102 103 3 201 202 203 ENQUEUE 101 ENQUEUE 201 ENQUEUE 102 ENQUEUE 202 ENQUEUE 103 ENQUEUE 203 DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 259001 259002 259003 259004 259005 6 260001 260002 260003 260004 260005 260006 ENQUEUE 259001 ENQUEUE 260001 ENQUEUE 259002 ENQUEUE 259003 ENQUEUE 259004 ENQUEUE 259005 DEQUEUE DEQUEUE ENQUEUE 260002 ENQUEUE 260003 DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0
Sample Output
Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001#include<iostream> #include<string> using namespace std; const int maxp=1000000; const int maxt=1000; const int maxn=200000+10; struct node { int p; int pre,next; }r[maxn]; int used; //队列新节点的下标 int belong[maxp]; //元素所在团队 int pos[maxt]; //团队i中最后一个最后一个元素在队列中的位置 int st,ed; //队首队尾指针 int main(void) { int t,loop=0; cin>>t; while(t) { for (int i=0;i<t;i++) { int m; cin>>m; for(int j=0;j<m;j++) { int x; cin>>x; belong[x]=i; } pos[i]=-1; } used=0; st=ed=-1; if(loop) //非第一测试用例输出空行 cout<<endl; cout<<"Scenario #"<<++loop<<endl; string s; cin>>s; while(s!="STOP") { if(s=="ENQUEUE") { int x; cin>>x; r[used].p=x; int s=pos[belong[x]]; if(s<0) //队列无此团队元素处理 { r[used].pre=ed; r[used].next=-1; if(ed>=0) r[ed].next=used; ed=used; //类似于双向链表 if(st<0) st=used; } else { r[used].pre=s; r[used].next=r[s].next; if(s==ed) //若是最后一个团队 ed=used; r[s].next= used; if(r[used].next>=0) r[r[used].next].pre=used; } pos[belong[x]]=used++; } else //处理输出 { cout<<r[st].p<<endl; if(st==pos[belong[r[st].p]]) pos[belong[r[st].p]]=-1; st=r[st].next; } cin>>s; } cin>>t; } return 0; }
这题花了好长时间，其实运用双向链表还是很容易的，指针的实现就是借助指向地址的变量来实现。这个代码理解队列，只能从逻辑上想，不能从存储结构上去考虑队列的排列
最后
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