HDU1387(队列+hash)

Team Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 979    Accepted Submission(s): 331

Problem 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.

 

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:

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.

 

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

23 101 102 1033 201 202 203ENQUEUE 101ENQUEUE 201ENQUEUE 102ENQUEUE 202ENQUEUE 103ENQUEUE 203DEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP25 259001 259002 259003 259004 2590056 260001 260002 260003 260004 260005 260006ENQUEUE 259001ENQUEUE 260001ENQUEUE 259002ENQUEUE 259003ENQUEUE 259004ENQUEUE 259005DEQUEUEDEQUEUEENQUEUE 260002ENQUEUE 260003DEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP0

 

Sample Output

Scenario #1101102103201202203Scenario #2259001259002259003259004259005260001
 
 
直接用队列+hash,模拟会由于查找插入位置而导致超时,时间复杂度也没有升高，超时原因真心不知道#include<iostream>#include<vector>using namespace std;#define MAX 999999+10int data[MAX];
int main(){ char str[20]; int n,i,j,m,temp; int tag=0; while(scanf("%d",&n)!=EOF&&n) {  tag++;  memset(data,0,sizeof(data));  vector<int>team;  vector<int>::iterator iter,index;  for(i=1;i<=n;i++)  {   scanf("%d",&m);   for(j=1;j<=m;j++)   {    scanf("%d",&temp);    data[temp]=i;   }  }    printf("Scenario #%d\n",tag);  while(scanf("%s",str)&&strcmp(str,"STOP")!=0)//&&str!="STOP")  {
  if(strcmp(str,"ENQUEUE")==0)  {  scanf("%d",&temp);  index=team.end();  for(iter=team.begin();iter!=team.end();iter++)  {  if(data[(*iter)]==data[temp])  {  index=iter+1;  break;  }  }  team.insert(index,temp);  }  else if(strcmp(str,"DEQUEUE")==0)  {  printf("%d\n",(*team.begin()));  team.erase(team.begin());  }  }  printf("\n");  team.clear(); } return 0;}
 
 
用队列模拟，具体根据所属团号作为队列数组下标，插入式时间复杂度为为O(1)，是在看了大神的代码后才自己敲出来的
#include<iostream>#include<queue>#include<map>using namespace std;
int main(){ char str[20]; int n,i,j,m,temp; int tag=0; bool visited[1010]; while(scanf("%d",&n)!=EOF&&n) {  tag++;  queue<int>que[1010],tol_que;  map<int,int>data;  memset(visited,0,sizeof(visited));  for(i=1;i<=n;i++)  {   scanf("%d",&m);   for(j=1;j<=m;j++)   {    scanf("%d",&temp);    data[temp]=i;
   }  }    printf("Scenario #%d\n",tag);  while(scanf("%s",str)&&strcmp(str,"STOP")!=0)//&&str!="STOP")  {      if(strcmp(str,"ENQUEUE")==0)   {    scanf("%d",&temp);    que[data[temp]].push(temp);    if(visited[data[temp]]==false)    {     tol_que.push(data[temp]);     visited[data[temp]]=true;    }   }   else if(strcmp(str,"DEQUEUE")==0)   {    printf("%d\n",que[tol_que.front()].front());    que[tol_que.front()].pop();    if(que[tol_que.front()].empty()==true)    {     visited[tol_que.front()]=false;     tol_que.pop();
    }
   }  }  printf("\n"); } return 0;}